This answer shows that all mentioned categories are not cartesian closed.
Here is an This answer for $\mathbf{Met}$ and $\mathbf{Met}_\infty$shows that all mentioned categories are not cartesian closed.
$\mathbf{Met}$ is not cartesian closed
Alternatively, andRemark. The same proof works in the category of pseudo-metric spaces.
$\mathbf{Met}_\infty$ is not cartesian closed
For this will then also work for $\mathbf{Met}_\infty$category, we continue the previous proof and check that the function $d$ on the non-expansive maps does not satisfy the triangle inequality. Without this condition, the category would indeed be cartesian closed. Here is an example*.
$\mathbf{Met}_c$ is not cartesian closed
Assume $\mathbf{Met}_c$ is cartesian closed. Notice that this category has all coproducts, in particular all copowers. Every cartesian closed category with copowers has all powers (proof). But $\mathbf{Met}_c$ does not have all powers. In fact, most uncountable products do not exist by MSE/139168.
$\mathbf{Haus}$ is not cartesian closed
It has been pointed out by Ben that actually MSE/1255678 can be used directly to settle $\mathbf{Haus}$. All the spaces are Hausdorff, the only one where it is not immediately clear is the $\mathbf{Top}$-colimit of the spaces $X_n \times \mathbb{Q}$. But this admits a continuous bijection to a Hausdorff space, and such a space is always Hausdorff. Thus, it is already the $\mathbf{Haus}$-colimit.
Some general thoughts
Here are some thoughts forthat apply to $\mathbf{Haus}$ (andand similar categories). The topology of a space is determined by the relation of net convergence. And this can be encoded via continuous maps, see Reference for convergent nets as continuous functions from $D\cup\{\infty\}$ for example. Specifically, a net indexed by a directed set $D$ in $[X,Y]$ (the potential Hom-object in $\mathbf{Haus}$) is just a map $D \to [X,Y]$, and it converges if it has a continuous extension to a suitably defined topological space $D \cup \{\infty\}$. We may assume w.l.o.g. that $D$ has no largest element (otherwise convergence is boring), but then this space is Hausdorff! Hence, continuous maps $D \cup \{\infty\} \to [X,Y]$ correspond to continuous maps $(D \cup \{\infty\}) \times X \to Y$. This means: If $(f_d)$ and $f$ are continuous maps $X \to Y$, we have $f_d \to f$ if and only if the map $$(D \cup \{\infty\}) \times X \to Y, \, (d,x) \mapsto f_d(x),\, (\infty,x) \mapsto f(x)$$ is continuous. The map is continuous at points $(d,x)$ anyway, and continuity at $(\infty,x)$ means:
So the only thing left to do is to check if the four axioms for net convergence (Kelley, General topology, Ch. 2, Thm. 9) are satisfied. Two axioms are easy: constant nets converge, and subnets of convergent nets converge to the same element. The other two are not that obvious (and cannot hold in general).
It has been pointed out by Ben that actually MSE/1255678 can be used directly to settle $\mathbf{Haus}$. All the spaces are Hausdorff, the only one where it is not immediately clear is the $\mathbf{Top}$-colimit of the spaces $X_n \times \mathbb{Q}$. But this admits a continuous bijection to a Hausdorff space, and such a space is always Hausdorff. Thus, it is already the $\mathbf{Haus}$-colimit.
Here is a short proof why $\mathbf{Met}_c$ is not cartesian closed. Assume it is. Notice that this category has all coproducts, in particular all copowers. Every cartesian closed category with copowers has all powers (proof). But $\mathbf{Met}_c$ does not have all powers. In fact, most uncountable products do not exist by MSE/139168. Incidentally, this proof has been generated by CatDat's deduction system a few days ago, hence it is also shown on the CatDat page for $\mathbf{Met}_c$, but I didn't notice before.