Let $X$ be a subset of $\textbf{R}$, let $E$ be a subset of $X$, $x_{0}$ be an adherent point of $E$, let $f:X\to\textbf{R}$ be a function, and let $L$ be a real number. Let $\delta > 0$. Then we have \begin{align*} \lim_{x\to x_{0};x\in E}f(x) = L \Longleftrightarrow \lim_{x\to x_{0};x\in E\cap(x_{0}-\delta,x_{0}+\delta)}f(x) = L \end{align*}
My solution
Let us prove the implication $(\Rightarrow)$ first.
According to the definition of limit, for every $\varepsilon > 0$, there is a $\eta > 0$ such that for every $x\in E$ one has that \begin{align*} |x - x_{0}| < \eta \Rightarrow |f(x) - L| < \varepsilon \end{align*}
Then for every $\varepsilon > 0$, there corresponds a $\zeta = \min\{\delta,\eta\} > 0$ such that for every $x\in E\cap(x_{0}-\delta,x_{0}+\delta)$ we have that \begin{align*} |x - x_{0}| < \zeta \Rightarrow |f(x) - L| < \varepsilon \end{align*} and we are done with the first part
Let us prove the implication $(\Leftarrow)$ now.
Once again, for every $\varepsilon > 0$, there is $\eta > 0$ such that for every $x\in E\cap(x_{0}-\delta,x_{0}+\delta)\subseteq E$, we have that \begin{align*} |x - x_{0}| < \eta \Rightarrow |f(x) - L| < \varepsilon \end{align*}
and the result holds because of the last inclusion.
Could someone please tell me if the wording of my proof is correct? If not, could you point out where is the flaw?