I have two examples using measurable cardinals, and two using real-valued measurable cardinals.
Example 1. In:
Mackey, George W., Equivalence of a problem in measure theory to a problem in the theory of vector lattices, Bull. Am. Math. Soc. 50, 719-722 (1944). ZBL0060.13402.
Mackey observed that if there exists a countably-additive 2-valued measure $\mu : \mathcal{P}(X) \rightarrow \{0,1\}$, then the mapping $f \mapsto \int_X f d\mu$ is a bounded linear map $\mathbb{R}^X \rightarrow \mathbb{R}$ that is not continuous for the product topology. Recall that a locally convex space $E$ is bornological iff every bounded linear map of locally convex spaces $E \rightarrow F$ is continuous (continuous always implies bounded). This shows that bornological spaces are not closed under product in the category of locally convex spaces if measurable cardinals exist. In fact, it is an iff (see Schaefer's Topological Vector Spaces, Chapter II, Exercise 19 (d)).
Example 2. A measurable space $(X,\Sigma)$ is said to be $\sigma$-perfect if the mapping from points $x \in X$ to two-valued measures $\Sigma \rightarrow 2$ provided by $x \mapsto \delta_x$ is a bijection. Being $\sigma$-perfect is the measurable-space analogue of being a sober space in topology (and $\Sigma$ separating the points of $X$ is both the analogue of being $T_0$ and Hausdorff, a disanalogy with the topological case where there are non-sober $T_0$ spaces but all Hausdorff spaces are sober).
Directly from the definition, the measurable space $(X,\mathcal{P}(X))$ is $\sigma$-perfect iff $|X|$ is less than the first measurable cardinal. It is also easily verified that the measurable space coproduct of a family $(X_i,\Sigma_i)_{i \in I}$ of non-empty $\sigma$-perfect measurable spaces is $\sigma$-perfect iff $|I|$ is less than the first measurable cardinal.
Example 3: It is well-known that every Borel probability measure on a Polish space $X$ is inner regular with respect to the compact subsets and $\tau$-additive (the latter can be proved by the former, or directly using the Lindelöfness of separable metric spaces). The question arises as to whether this can be proved for all completely metrizable spaces, of arbitrarily large density character (the smallest cardinality of a dense subset).
If $X$ is a set admitting a probability measure $\mu : \mathcal{P}(X) \rightarrow [0,1]$ vanishing on singletons, we can make it into a metrizable space with the discrete metric (and it has density character $|X|$). Since the compact subsets are the finite subsets, this measure is not Radon. Likewise the measure is not $\tau$-additive, because
$$
\sum_{x \in X} \mu(\{x\}) = 0 \neq 1 = \mu(X) = \mu\left(\bigcup_{x \in X}\{x\}\right),
$$
using the fact that singletons are open. So real-valued measurable cardinals provide counterexamples to the attempt to extend these properties to arbitrary metrizable spaces. In fact, we have that for a metrizable space $X$, all probability measures are $\tau$-additive iff the density character of $X$ is below the first real-valued measurable, and for a completely metrizable space $X$, this is equivalent to all probability measures being inner regular with respect to the compact sets.
For a proof see Fremlin's Measure Theory 438J (c) for the fact about $\tau$-additivity for a metrizable space and 438H for inner regularity in the case of a completely metrizable space. (In each case to get to the statement I used you need the fact that the weight (minimal cardinality of a base) and density character are the same for a metrizable space.) These results are also contained in some editions of Billingsley's Convergence of Probability Measures, but not others.
It's should be mentioned that since $\mathfrak{c}$ can be real-valued measurable, these things can fail for $\ell^\infty(\mathbb{N})$, $L^\infty([0,1])$ or $B(\ell^2)$ in the norm topology, i.e. they can have non-Radon probability measures.
Example 4. Consider two measure spaces $(X,\Sigma_X,\mu_X)$ and $(Y,\Sigma_Y,\mu_Y)$ such that $L^\infty(X)$ and $L^\infty(Y)$ are W$^*$-algebras (equivalently, von Neumann algebras when represented in $L^2$). Such measure spaces are called localizable (sometimes an equivalent condition is used as the definition). Given a null-set reflecting measurable map $f : X \rightarrow Y$, we always get a $\sigma$-normal *-homomorphism $L^\infty(f) : L^\infty(Y) \rightarrow L^\infty(X)$, where $\sigma$-normal means that for every countable disjoint family of projections $(p_i)_{i \in I}$ in $L^\infty(Y)$ we have $$
L^\infty(f)\left(\bigvee_{i \in I}p_i\right) = \bigvee_{i \in I} L^\infty(f)(p_i).
$$
We might wonder if it's always a normal *-homomorphism, where the analogous fact holds for disjoint families of projections of arbitrary cardinality (often authors restrict to $\sigma$-finite measures to simplify this away). Well, if $X$ admits a probability measure $\mu : \mathcal{P}(X) \rightarrow [0,1]$ vanishing on singletons, the answer is no. We just take $\nu$ to be the counting measure on $X$, whose only null-set is $\emptyset$, and then $\mathrm{id}_X : (X,\mathcal{P}(X),\mu) \rightarrow (X,\mathcal{P}(X),\nu)$ is null-set reflecting and gives us a non-normal map, as the join of the family of of projections $(\chi_{\{x\}})_{x \in X}$ in $L^\infty(X,\nu) = \ell^\infty(X)$ is not preserved by $L^\infty(\mathrm{id}_X)$.
In fact, every $\sigma$-normal *-homomorphism between commutative W$^*$-algebras is normal iff there are no real-valued measurable cardinals. As far as I know this was first shown by Fremlin in his chapter on Measure Algebras in Volume III of the Handbook of Boolean Algebras, Remark 4.13 (b).
This example is the cause of this answer existing, as my independent re-discovery of Fremlin's argument, for commutative W$^*$-algebras instead of measure algebras, led to me learning what a measurable cardinal was at all. I originally made a counterexample under the assumption that all subsets of $\mathbb{R}$ were Lebesgue measurable, and then tried to find a suitable alteration of that assumption that would be compatible with the axiom of choice, which led to measurable cardinals.
I will also mention that the first place the mistake mentioned in the question occurred was as early as possible, in Hewitt's paper:
Hewitt, Edwin, Rings of real-valued continuous functions. I, Trans. Am. Math. Soc. 64, 45-99 (1948). ZBL0032.28603.
It is stated that every discrete space is realcompact in Theorem 52 on p. 87 (p. 43 of the pdf). The implicit assumption of no measurable cardinals occurs in the "proof" that $\bigcap_{n = 0}^\infty C_n = \emptyset$.
This error is not mentioned by Dieudonné in his famous review* that points out many of the errors - but of course he couldn't provide a counterexample.
*This should be visible to MathSciNet subscribers.
