Proving that a function $f$ converges to a number $L$ if and only if the sequence $(f(a_n))_{n=m}^\infty$ converges to $L$.

This is an exercise that I am trying to prove. I am pretty sure that the proof of the first implication is correct. I am not so sure about the second one. I would appreciate if someone could check it for me.

Theorem: Let $X \subset R$ and let $f : X \to R$ be a function. Let $E \subset X$, let $x_0$ be an adherent point of $E$ and let $L \in R$. Then the function $f$ converges to $L$ at $x_0$ in $E$ if and only if for every sequence $(a_n)_{n=m}^\infty$ consisting entirely of elements of $E$ that converges to $x_0$, the sequence $(f(a_n))_{n=m}^\infty$ converges to $L$.

Proof: First suppose that $\lim_{x\to x_0; x\in E}f(x) = L$. Then for every $\varepsilon > 0$, there exists a $\delta > 0$ such that for all those values of $x\in E$ for which $|x- x_0|< \delta$, we have that $|f(x) - L| < \varepsilon$.

Let $\varepsilon > 0$. Let $\varepsilon ' = \delta$. Given that $(a_n)_{n=m}^\infty$ converges to $x_0$, there exist an $N\ge m$ such that for all $n > N$, we have that $|a_n - x_0| < \varepsilon' = \delta$. Because every element of the sequence is from $E$, we have that $|f(a_n) - L| < \varepsilon$. Thus, for every $\varepsilon > 0$, there exists an $N\ge m$ such that for all $n > N$, we have that $|f(a_n) - L| < \varepsilon$ and thus the sequence $(f(a_n))_{n=m}^\infty$ converges to $L$.

Conversely suppose that the sequence $(f(a_n))_{n=m}^\infty$ converges to $L$. We are also given that the sequence $(a_n)_{n=m}^\infty$ converges to $x_0$. Thus for every $\varepsilon > 0$ there exists an $N\ge m$ such that for all $n > N$, we have that $|f(a_n) - L| < \varepsilon$ and $|a_n - x_0|<\varepsilon$.

We have to show that for every $\varepsilon > 0$, there exists a $\delta > 0$ such that for all those values of $x\in E$ for which $|x - x_0| < \delta$, we have that $|f(x) - L| < \varepsilon$.

Let $\varepsilon > 0$ and $0 < \varepsilon '< \varepsilon$. Then there exists an $N\ge m$ such that for all $n > N$, we have that $|a_n - x_0| < \varepsilon '$ and $|f(a_n) - L| < \varepsilon '$. Let $\delta = \varepsilon '$. Then for all those values of $x\in E$ for which $|x - x_0| < \delta$, we have that $|f(x) - L| < \varepsilon$. If some element $x$ of $E$ is not in the sequence, then we can add it to the sequence and the sequence would still converge to $x_0$ and the above will hold.

I am not too sure if the proof of the second direction is correct or not. Could someone please check if it is correct or if it can be improved and written in a better way?