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

Rather than making a new post, I decided to edit this one. I tried proving the contrapositive and I feel like I have it right now. This is also suggested in the comments. I would appreciate if someone could check if this is correct or not.

Theorem: Let $X \subset R$, 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 following are equivalent.

$(a): $$f$ converges to $L$ at $x_0$ in $E$.

$(b): $ For every sequence $(a_n)_{n=m}^\infty$ which consists entirely of elements from $E$ and converges to $x_0$, the sequence $(f(a_n))_{n=m}^\infty$ converges to $L$.

I have already proven $a \implies b$. I am now trying to prove that $b \implies a$. I will be proving the contrapositive $\neg a \implies \neg b$.

Proof: We will prove the contrapositive. Suppose that $f$ does not converge to $L$ at $x_0$ in $E$. Then there exists an $\varepsilon > 0$ such that for all $\delta > 0$, there exists an $x\in E$ for which $|x - x_0| < \delta$ implies $|f(x) - L| \ge \varepsilon$.

Suppose for the sake of contradiction that $b$ is true. Let $(a_n)_{n=m}^\infty$ be a sequence that converges to $x_0$ and also the sequence $(f(a_n))_{n=m}^\infty$ converges to $L$.

Let $\delta > 0$. Then there exists an $N\ge m$ such that for all $n > N$, we have that $|a_n - x_0| < \delta$.

Let $\varepsilon > 0$. Then there exists an $N'> m$ such that for all $n'> N'$, $|f(a_n) - L| < \varepsilon$.

Let $N'' = max(N, N')$. Then for all $n > N''$, if $|a_n - x_0| < \delta$, then $|f(a_n) - L| < \varepsilon$.

But we know that there exists an $\varepsilon > 0$ such that for all $\delta > 0$, there exists an $x\in E$ such that $|x - x_0| < \delta$ implies $|f(x) - L| \ge \varepsilon$. If $x$ is not in the sequence $(a_n)_{n=m}^\infty$, we can add it to the sequence and we would have a sequence that converges to $x_0$ and also a sequence $(f(a_n))_{n=m}^\infty$ that converges to $L$. Since $|x - x_0| < \delta$, we should have that $|f(x) - L| < varepsilon$. But this contradicts our hypothesis that $|f(x) - L| \ge \varepsilon$. Thus, we have that $\neg b \implies \neg a$ and $b \implies a$.