Let $X$ be a subset of $\textbf{R}$, let $f:X\rightarrow\textbf{R}$ be a function, let $E$ be a subset of $X$, let $x_{0}$ be an adherent point of $E$, and let $L$ be a real number. Then the following two statements are logically equivalent
(a) $f$ converges to $L$ at $x_{0}$ in $E$.
(b) For every sequence $(a_{n})_{n=0}^{\infty}$ which consists of elements of $E$ and converges to $x_{0}$, the sequence $(f(a_{n}))_{n=0}^{\infty}$ converges to $L$.
MY ATTEMPT
(a) If $f$ converges to $L$ at $x_{0}$, for every $\varepsilon > 0$, there exists an $\delta > 0$ such that \begin{align*} |x - x_{0}| < \delta \Longrightarrow |f(x) - L| \leq \varepsilon \end{align*}
Since the sequence $(a_{n})_{n=0}^{\infty}$ converges to $x_{0}$, for every $\delta > 0$, there exist a natural number $N\geq 0$ such that \begin{align*} n\geq N \Longrightarrow |a_{n} - x_{0}| \leq \delta \end{align*}
Therefore, if we make the substitution $x = a_{n}$, for every $\varepsilon > 0$, there exists a natural number $N\geq 0$ such that \begin{align*} n\geq N \Longrightarrow |a_{n} - x_{0}| \leq \delta \Longrightarrow |f(a_{n}) - L| \leq \varepsilon \end{align*} whence we conclude that $f(a_{n})\to L$, just as desired.
(b) Conversely, let us assume that $a_{n}\to x_{0}$ and $f(a_{n})\to L$.
We have to prove that, for every $\varepsilon > 0$, there exists a positive real number $\delta > 0$ such that \begin{align*} |x - x_{0}| < \delta \Longrightarrow |f(x) - L| \leq \varepsilon \end{align*}
Then I get stuck. Can someone help me to finish the proof?