Let $(X,d_{X})$ and $(Y,d_{Y})$ be metric spaces, with $Y$ complete, and let $E$ be a subset of $X$. Let $f_{n}$ be a sequence of functions from $E$ to $Y$, and suppose that this sequence converges uniformly in $E$ to some function $f:E\to Y$. Let $x_{0}\in X$ be an adherent point of $E$, and suppose that for each $n$ the limit $\lim_{x\to x_{0};x\in E}f_{n}(x)$ exists. Then the limit $\lim_{x\to x_{0};x\in E}f(x)$ also exists and we have that \begin{align*} \lim_{n\rightarrow\infty}\lim_{x\rightarrow x_{0};x\in E}f_{n}(x) = \lim_{x\rightarrow x_{0};x\in E}\lim_{n\rightarrow\infty}f_{n}(x) \end{align*}
MY ATTEMPT
Let us reinforce the definitions involved.
Since $f_{n}\to f$ uniformly, let $\varepsilon > 0$. Then there corresponds a $N\geq 0$ such that for every $x\in E$ we have that \begin{align*} n\geq N \Rightarrow d_{Y}(f_{n}(x),f(x)) < \varepsilon \end{align*}
In accordance to the RHS, let $\varepsilon > 0$. Then there corresponds a $\delta > 0$ such that for every $x\in E$ we have that \begin{align*} d_{X}(x,x_{0}) < \delta \Rightarrow d_{Y}(f(x),L) < \varepsilon \end{align*}
Similarly, accoring to the LHS, let $\varepsilon > 0$. Then there corresponds a natural $M \geq 0$ such that \begin{align*} n\geq M \Rightarrow d_{Y}(L_{n},L') < \varepsilon \end{align*}
We have to prove that $L = L'$. Let us consider otherwise that $L\neq L'$.
But then I get stuck. Can someone please help me to prove it?