Subsequence converging to supremum of sequence 
Let $(a_n)$ be a bounded sequence of reals satisfying
  $$a_n<L:=\sup\{a_n\mid n\in\mathbb{N}\}$$for all $n\in\mathbb{N}$. Prove that there exists a subsequence $a_{n_j}$ that converges to $L$.

I was thinking of constructing a sequence such that $L-\frac{1}{n}<a_{n_j}<L$, but how do we know that there exist elements of $a_n$ such that $L-1\in a_n$ for example?
There is a theorem in my analysis textbook that says:
A sequence $(s_n)$ has a subsequence converging to $t$ if and only if the set $\{n\in\mathbb{N}\mid |a_n-t|<\varepsilon\}$ is infinite for all $\varepsilon>0$. 
Could someone provide a proof using this theorem?
 A: No $M<L$ is an upper bound for the set $A=\{a_n:n\in \Bbb N\}.$ 
Let $n_1=1.$  For $j\in \Bbb N,$ define $n_{j+1}$ recursively as follows: 
For brevity let $V(j)=\max \{a_k:1\leq k\leq n_j\}.$ We have $V(j)<L.$ 
Let $M(j)=\max (L-2^{-j},V(j)).$ Then $M(j)<L$ so $M(j)$ is not an upper bound for $A.$ So let $n_{j+1}$ be the least (or any) $n$ such that $a_n>M(j).$ 
We have $n_{j+1}>n_j$ because  if $k\leq n_j$ we have $a_k\leq V(j)\leq M(j)<a_{n_{j+1}},$ implying $k\ne n_{j+1}.$  
Conclusion: The sequence $(a_{n_j})_{j\in \Bbb N}$ converges to $L$ because we have $L-2^{-j}\leq M(j)< a_{n_{j+1}}<L.$
Remark: The reason this will not work if we replaced $L$ with some $L'>L$ is that for some (large enough) $j$ we would have $M(j)\geq L'-2^{-j}>\sup A,$ whereupon $n_{j+1}$ won't exist.  
A: This is quite a few years on, but I love this question and have been pondering over it a while. 
Let the sequence be ${(a_n)}_{n=m}^{∞}$
Now $L:=sup\{a_n| n \in \mathbb N\}$.
Take any $\epsilon>0$ which is real. 
Then, by the definition of a supremum, there will exist an $n$ such that $a_n>L-\epsilon$.
Let this $n=n^*$
For this very $n^*$, $a_{n^*}<L+\epsilon$.
Therefore for that $n^*$, $|a_{n^*} - L|<\epsilon$. 
Now, suppose that for an N≥m, the sequence ${(a_n)}_{n=N}^{∞}$ has no term that is greater than $L-\epsilon$. Then, the sequence ${(a_n)}_{n=m}^{N}$ is finite and has a term which is equal to the least upper bound or supremum. But this contradicts the fact that the supremum is strictly greater than $a_n$ for all $n$. Hence, for every $N≥m$, you can always find an n such that $a_n>L-\epsilon$. 
This proves, for every $N≥m$ and every real $\epsilon >0$, there exists a $n≥N$, such that  $|a_n - L|<\epsilon$.(1)
Now we know, "a sequence $(s_n)$ has a subsequence converging to $t$ if and only if the set $\{n\in \mathbb N∣ |a_n−t|<\epsilon\}$ is infinite for all $\epsilon>0$." [This is the defintion of a limit point]
By (1), this set will be infinite for this sequence and thus, there will exist a subsequence which converges to $L$. 
