If $\limsup\limits_{n\rightarrow \infty} a_n=a< \infty$, then there exists for all $\varepsilon >0$ a $N\in \mathbb{N}$ such that $a_n\leq a+\varepsilon$ for all $n\in \mathbb{N}$, $n\geq N$
My attempt:
Suppose there are infinite elements $a_{n_1},a_{n_2}, a_{n_3},…$ with $a_{n_k}\geq a$. The sequence $(a_{n_k})_k$ is bounded above, otherwise $(a_n)_n$ wouldn't be bounded and $\limsup\limits_{n\rightarrow \infty} a_n=+\infty$. Hence $(a_{n_k})_k$ has – according to Weierstraß and Bolzano – a convergent subsequence $(a_{n_{k_j}})$ with a limit $\geq a$} since $(a_{n_{k_j}})\geq a$ for every $j\in \mathbb{N}$.
Best Answer
You don't need to go as far as using Bolzano-Weierstraß theorem. We can prove your statement directly.
Recall that $\limsup\limits_{n\rightarrow \infty} a_n = \lim\limits_{n\to\infty} \sup\limits_{k\geq n} a_k$. Let us fix $\epsilon > 0$. By definition of the limit, we know that there is some $N\in \mathbb N$ such that $\sup\limits_{k\geq N} a_k < a+\epsilon$. Thus, by definition of the supremum, we conclude that $a_k < a + \epsilon$ for every $k\geq N$, which is the conclusion you wanted.