if $\lim \sup a_n = l \in \mathbb R$, then given $\epsilon > 0, \exists N$ such as $a_n < l + \epsilon, \forall n \geq N$

You can do it by contraposition :

Suppose that $\forall N, \exists n>N, a_n \geq l+\epsilon$

You can construct by induction a sequence of indices $(n_i)$ this way :

$\exists n_0>0, a_{n_0} \geq l+\epsilon$

Suppose $n_{i}$ constructed, then by hypothesis, $\exists n_{i+1} > n_i , a_{n_{i+1}} \geq l+\epsilon$

A reccurence gives you the existence of a subsequence $a_{n_i}$ of $a_n$ such that $\forall i, a_{n_i} \geq l+\epsilon$

Now, because $n_i \geq n$, you have $\sup_{n>N} a_n \geq \sup_{i>N} a_{n_i} \geq l+\epsilon$

So $\limsup a_n \geq l+\epsilon$

Hence you proved that

$\forall N, \exists n>N, a_n \geq l+\epsilon$ imply $\limsup a_n \geq l+\epsilon$

or, by contraposition,

$\limsup a_n < l+\epsilon $ imply $\exists N, \forall n > N, a_n < l+\epsilon$

Recall the definition of limsup: $$ \limsup_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}\sup_{m\geq n}a_{n}. $$ Hint: in order to arrive at a contradiction, suppose there exists some $\epsilon>0$ such that for all $N\geq1$, there exists an $n\geq N$ with $a_{n}\geq\ell+\epsilon$. It follows that we can pick a subsequence $(a_{n_{i}})_{i}$ satisfying $a_{n_{i}}\geq\ell+\epsilon$...

Prove by contradiction: If $$\exists \epsilon>0 \ \forall N \exists \ n>N \ {a_n>l+\epsilon}$$ then we can find subsequence $(a_{n_{k}})$ that $\forall k \ (a_{n_{k}})>l+\epsilon $ We get a contradiction: $$ l+\epsilon< \limsup a_{n_{k}} \leq \limsup a_n =l $$ Q.E.D

($\limsup$ is the bigest partial limit, or by the difenition with the partial $\sup$)

You can do this directly:

$y_n=\sup_{ n\leq k } \left \{ a_k \right \}$ and $y_n\searrow l$.

There is an $N\in \mathbb N$ s.t. $n>N\Rightarrow y_n<l+\epsilon$.

Therefore, if $n>N$, we have $a_n\leq \sup_{ n\leq k } \left \{ a_k \right \}=y_n<l+\epsilon$