Let $(a_{n})_{n=0}^{\infty}$ be a sequence of real numbers, and let $L$ be a real number. Then the following two statements are logically equivalent.
(a) $L$ is a limit point of $(a_{n})_{n=0}^{\infty}$
(b) There exists a subsequence of $(a_{n})_{n=0}^{\infty}$ which converges to $L$.
My solution
Please consult here.
Firstly, as @zipirovich mentions in the comments under your question, in the first part, your statement of what you need to prove is a bit off. You say:
We need to prove that for every $\varepsilon > 0$, there is a strictly increasing function $f \colon \mathbf{N} \to \mathbf{N}$ such that $b_n = a_{f(n)} \to L$.
but nowhere is that $\varepsilon$ used in the latter half. It is actually hidden in the symbols "$b_n = a_{f(n)} \to L$". This is poor form—it would be much better to expand out the "$\to$" to show where the $\varepsilon$-dependence arises. You must have something like
. . .for every $\varepsilon > 0$. . .there exists $k \in \mathbf{N}$ such that. . .for all $n \geq k$. . .
Can you fill in the right form?
Secondly, the quoted statement is not what you want to prove. You want to prove that a convergent subsequence exists, or in other words:
We want to show that there is a strictly increasing function $f \colon \mathbf{N} \to \mathbf{N}$ such that for every $\varepsilon > 0$. . .
Note the difference made by swapping the order of the "for every" and "there exists" terms. You do go on to show what you claimed you need to show, but this does not help you prove that there is a convergent subsequence.
Just to elaborate, what you have shown is something like the following: if I give you an $\varepsilon > 0$, say $\varepsilon = 1$, then you're able to give me a subsequence all of whose terms are within distance $1$ of $L$. But, this does not mean that the subsequence gets arbitrarily close to $L$! It's possible that they are all within distance $1$, but also a distance $1/2$ away from $L$.
Next, if I give you $\epsilon = 1/2$, then you can give me some subsequence all of whose terms are within distance $1/2$ of $L$. But, this again does not mean that this subsequence gets arbitrarily close to $L$. Note that from your proof this subsequence has got nothing to do with the one you found when $\epsilon = 1$, which is partly the problem here.
For any fixed $\varepsilon > 0$, you are able to produce a subsequence which lies within the ball of radius $\epsilon$ centered at $L$. But, you have not yet produced one subsequence which gets arbitrarily close to $L$.
To construct the required subsequence, you need to follow something like this outline:
For each $k \in \mathbf{N}$, there exists $n_k \in \mathbf{N}$ such that $\lvert a_{n_k} - L \rvert < 1/k$. Moreover, we can ensure that $(n_{k})$ is a strictly increasing sequence of integers (why?). Hence, $(a_{n_k})$ is a convergent subsequence of $(a_n)$ converging to $L$.
Your second part is accurate (modulo some typos), although a tad confusing, since the roles of $n$ and $N$ appear to be interchanged.
Perhaps modifying the definition of convergence to use another alphabet would be helpful:
Let $(a_{f(n)})$ be a subsequence of $(a_n)$ that converges to $L$, where $f \colon \mathbf{N} \to \mathbf{N}$ is an increasing function. According to the definition of convergence, for every $\varepsilon > 0$, there is a natural number $M\geq 0$ such that for all $n \geq M$, we have \begin{align*} \lvert a_{f(n)} - L \rvert \leq \varepsilon. \end{align*}
Let $\varepsilon > 0$. If $0\leq N \leq M$, we can choose $n = f(M)$ as the natural number satisfying the desired property $|a_{n} - L|\leq\varepsilon$.
If $N = M+1$, then we can choose $n = f(M+1)$ as the natural number satisfying the desired property: $|a_{n} - L| \leq \varepsilon$.
If $N = M + k$, then we can choose $n = f(M+k)$ as the natural number satisfying the desired property: $|a_{n} - L|\leq\varepsilon$.
Since $\epsilon > 0$ was arbitrary, we have then proved that for every $\varepsilon > 0$ and every natural number $N\geq 0$ there is a natural number $n\geq N$ such that $|a_{\color{red}{n}} - L|\leq\varepsilon$.