Show that the sequence converges I am hoping to have my proof reviewed for verification. Thanks. 
Problem:

Let $X$ be a topological space and let $Y$ be a metric space. Let $f_n: X \to Y$ be a sequence of continuous functions. Let $x_n$ be a sequence of points in $X$ converging to $x \in X$. Show that if the sequence $(f_n)$ converges uniformly to $f$, then $f_n(x_n)$ converges to $f(x)$. 

Proof:
Since each $f_n$ is continuous, $Y$ is a metric space, and $f_n$ converges to $f$ uniformly, then $f$ is continuous.
Let $U$ be an open set containing $f(x)$. 
Take $V$ to be the open ball of some radius $ε$ centered at $f(x)$, where $V$ is contained within $U$. 
Let $W$ be the open ball of radius $\dfracε2$, centered at $f(x)$. Then so far we have: $W \subset V \subset U$. 
Then $f^{-1}(W)$ is an open set containing $x$. Then there exists some $n_1$ such that for $n > n_1, x_n \in f^{-1}(W)$. 
Then $f(x_n) \in W$ for $n > n_1$.
Then choose an $n_2$ such that $d(f(x), f_n(x)) < \dfracε2, \forall x, \forall n > n_2$. Such an $n_2$ exists since $f_n$ converges uniformly to $f$. (Note: $d$ is the distance function on the metric space $Y$). 
Take $n_3$ to be maximum of $n_1$ and $n_2$. Then for $n > n_3$, the sequence $f_n(x_n)$ is contained within $V$ and hence, $U$. So the sequence converges to $f(x)$ since $U$ was chosen to be an arbitrary open set containing $f(x)$.
 A: It's a completely valid proof which I would write differently (more formulae, fewer words): 
Take $\varepsilon > 0$. As $f$ is continuous at $x$ (for the reasons you stated) we can find an open neighbourhood $U$ of $x$ such that 
$$\text{ (1): } \forall p \in U : d(f(p) ,f(x)) < {\varepsilon \over 2}$$
And as $x_n \to x$, we can find $N_1 \in \mathbb{N}$ such that:
$$\text{ (2): } \forall n \ge N_1: x_n \in U$$
Combining (2) with (1) we thus have:
$$\text{ (2a): } \forall n \ge N_1: d(f(x_n), f(x)) < {\varepsilon \over 2}$$
And by uniform convergence of $f_n$ to $f$, we can find $N_2 \in \mathbb{N}$ such that:
$$\text{ (3): } \forall n \ge N_2: \forall x \in X: d(f_n(x), f(x)) < {\varepsilon \over 2}$$ 
Now taking $n \ge N_3 = \max(N_1, N_2)$, by the triangle inequality we see:
$$d(f_n(x_n), f(x)) \le d(f_n(x_n), f_n(x)) + d(f_n(x), f(x)) < {\varepsilon \over 2} + {\varepsilon \over 2} = \varepsilon$$
because we can apply (2a) and (3) at the end. As $\varepsilon>0$ was arbitrary, we have shown $f_n(x_n) \to f(x)$, as required.
