Limit points and subsequences I'm having trouble with proving this exercise:
Let $(a_{n})_{n = 0}^\infty$ be a sequence of real numbers, and $L \in \mathbb{R}$ . Prove:  
$L$ is a limit point of $(a_{n})_{n = 0}^\infty \Leftrightarrow$ There is a subsequence of $(a_{n})_{n = 0}^\infty$ that converges to L.
So far, I've done this:
$\Leftarrow:$
There is a subsequence of $(a_{n})_{n = 0}^\infty$ that converges to L. Assume $(b_{n})_{n = 0}^\infty$ is that subsequence, and $\epsilon \in \mathbb{R}$
$\Rightarrow \exists$ $n \in \mathbb{N}$ such that $|b_{n} - L| < \epsilon$ $\forall$ $n \geqslant N, \epsilon > 0$
$\Rightarrow$ $(b_{n})_{n = 0}^\infty$ is a subsequence of $(a_{n})_{n = 0}^\infty$, so $b_{n} = a_{m}$ for a certain $m \geqslant 0$
$\Rightarrow$ $\exists$ $m \in \mathbb{N}$ such that $|a_{m} - L| < \epsilon$ $\forall$ $m \geqslant N, \epsilon > 0$
$\Rightarrow L$ is a limit point of $(a_{n})_{n = 0}^\infty$
Is my left implication correct? And if not, how does it have to be?
Could you explain me the right implication? 
Thanks in advance!
 A: '⇐'
We want to show that L is a limit point of the sequence $({a}_{n})_{n \in\mathbb{N}}$.
This is equivalent to: $\forall \epsilon >0, \exists N \in \mathbb{N}: |{a}_{N}-L|< \epsilon$.
Proof: 
Let $\epsilon >0$. Consider the subsequence $({b}_{n})_{n \in\mathbb{N}}$. We know that $({b}_{n})$ converges to L. So $\exists N \in \mathbb{N}: |{b}_{N} -L|<\epsilon$.
Now ${b}_{N}={a}_{m}$ for some $m$. Then: $|{a}_{m} -L|<\epsilon$. This ends the proof.
'⇒'
We want to show that if $L$ is a limit point of $({a}_{n})_{n \in\mathbb{N}}$, then there exists a subsequence $({b}_{n})_{n \in\mathbb{N}}$ that converges to $L$.
Proof:
We know by definition of limit point:  $\forall \epsilon >0, \exists N \in \mathbb{N}: |{a}_{N}-L|< \epsilon$. Take $\epsilon $ of the form $1/n$ for $n>0$. Then we know:
 $\forall N >0, \forall p \in \mathbb{N}, \exists M>p \in \mathbb{N}: |{a}_{M}-L|< 1/N$.
This $M$ is dependent of $p$ and $N$ so we denote $M$ by $M(p,N)$ 
 Take $$ m_{0} = 0 \: ,\: m_{1} = M(m_{0},1) \: ,\: m_{2} = M(m_{1},2) \: , \: ...$$
We see directly that ${\left( m_{i} \right) }_{i \in \mathbb{N}}$ is an increasing sequence in $\mathbb{N}$.
Now define ${b}_{n}={a}_{m_{n}}$ for every $n \in \mathbb{N}$. It is easely seen that $({b}_{n})_{n \in\mathbb{N}}$ converges to L.
