Show $f_n(x_n) \to f(x)$ if $f_n \to f$ uniformly and $x_n \to x$. [duplicate]

Let $(X,d)$ be a metric space, let $f_n: X \to \mathbb R$ be a sequence of continuous functions such that $f_n \to f$ uniformly on $X$ and let $x_n$ be a sequence of points of $X$ with $x_n \to x$. Show $f_n(x_n) \to f(x)$.

This is my attempt, just wondering if it's correct.

Let $\varepsilon>0$ be given, then there exists $\delta>0$ such that $|x-y| < \delta$ implies $|f_n(x) - f_n(y)| < \varepsilon/2$.

As $x_n \to x$ in $X$, there exist some integer $N_1$ such that $|x_n - x| < \varepsilon_1$ for all $n \geq N_1$.

Let $\varepsilon_1 = \delta$. Then $|x_n - x| < \varepsilon_1 = \delta$ implies $|f_n(x_n) - f_n(x)| < \varepsilon/2$ for all $n ≥ N_1$.

As $f_n \to f$, there exists some integer $N_2$ such that $|f_n(x) - f(x)| < \varepsilon_2$ for all $n ≥ N_2$.

Let $\varepsilon_2 = \varepsilon/2$. Then $|f_n(x) - f(x)| < \varepsilon/2$ for all $n ≥ N_2$.

Let $N = \max\{ N_1 , N_2 \}$. Then $$|f_n(x_n) - f(x)| ≤ |f_n(x_n) - f_n(x)|+|f_n(x) - f(x)| < \varepsilon/2+\varepsilon/2=\varepsilon$$ for all $n ≥ N$.

marked as duplicate by Najib Idrissi, hardmath, Pierre-Guy Plamondon, Claude Leibovici, N. F. TaussigFeb 18 '16 at 12:24

First, a small technical detail. Notice that $X$ is a metric space, not a normed space, so probably $\vert x-y\vert$ is not defined there. What is defined is $d(x,y)$
Second. I think a better strategy, which uses $f_n\rightarrow f$ uniformly, is this: $$\vert f_n(x_n)-f(x)\vert \leq \vert f_n(x_n)-f(x_n)\vert +\vert f(x_n)-f(x)\vert$$ The first term in the RHS becomes very small because $f_n\rightarrow f$ uniformly. The second term in the RHS becomes very small because $f$ is continuous and $x_n\rightarrow x$
Let $\varepsilon>0$ be given, then there exists $\delta>0$ such that $|x-y| < \delta$ implies $|f_n(x) - f_n(y)| < \varepsilon/2$.
This is true indeed for each $n$ individually. For each $\varepsilon>0$ and for each $n$ there is $\delta_n$ such that... . The problem is that you don't know if for each $\varepsilon>0$ there is a $\delta$ such that "for all $n$" ...