Proving that a sequence converges to L Given a sequence $(a_{n})_{n=1}^{\infty}$ that is bounded. Let $L \in R$. Suppose that for every subsequence $(a_{n{_{k}}})_{k=1}^{\infty}$ , either $$\lim_{k \to \infty}a_{n{_{k}}} = L$$
or $(a_{n{_{k}}})_{k=1}^{\infty}$ does not converge. Prove that $(a_{n})_{n=1}^{\infty}$ converges to L.
I am trying to prove the problem above. I need to prove that the following sequence converges to $L$. I tried to prove this by contradiction, where I assume that $(a_{n})_{n=1}^{\infty}$ does not converge to $L$, but I dont seem to be able to know how to continue. I am aware that from  Bolzano-Weierstrass Theorem, every bounded sequence has a convergent subsequence, so from that is it safe to say that $(a_{n{_{k}}})_{k=1}^{\infty}$ does not converge is false, and that the subsequence does indeed converge to $L$ ?
Could somebody provide me with a sufficient proof for this problem ? A proof by contradiction will be ideal. Thanks for the help.
 A: @Alex Francisco has provided a proof by contradiction. Here's a direct proof:
Since $(a_{n})$ is bounded, then $-\infty<\liminf_{n}a_{n}\leq\limsup_{n}a_{n}<\infty$. And there are subsequences $(a_{n_{k}})$ and $(a_{m_{l}})$ such that $a_{n_{k}}\rightarrow\limsup_{n}a_{n}$ and $a_{m_{l}}\rightarrow\liminf_{n}a_{n}$. Such two subsequences are convergent, so by assumption we have both $a_{n_{k}}\rightarrow L$ and $a_{m_{l}}\rightarrow L$.
Therefore, $\liminf_{n}a_{n}=\limsup_{n}a_{n}=L$, in other words, $\lim_{n}a_{n}=L$.
A: You are almost there. Suppose $a_n \not\to L \ (n \to \infty)$, then by the $ε$-$N$ definition of limit, there exists $ε > 0$ and a subsequence $\{a_{n_k}\}$ such that$$
|a_{n_k} - L| \geqslant ε. \quad \forall k \in \mathbb{N}_+ \tag{1}
$$
Note that $\{a_{n_k}\}$ is bounded as $\{a_n\}$ is, thus there exists a convergent subsequence $\{a_{n_{k_l}}\}$ of $\{a_{n_k}\}$, and from the given condition there is$$
\lim_{l \to \infty} a_{n_{k_l}} = L,
$$
which is contradictory to (1).
