Here is Prob. 8, Sec. 21, in the book Topology by James R. Munkres, 2nd edition:
Let $X$ be a topological space and let $Y$ be a metric space. Let $f_n \colon X \to Y$ be a sequence of continuous functions. Let $x_n$ be a sequence of points of $X$ converging to $x$. Show that if the sequence $\left( f_n \right)$ converges uniformly to $f$, then $\left( \ f_n \left( x_n \right) \ \right)$ converges to $f(x)$.
My Attempt:
As $f_n \colon X \to Y$ is a sequence of continuous functions converging uniformly to the function $f \colon X \to Y$, so $f$ is also continuous, by Theorem 21.6 in Munkres.
Let $\varepsilon > 0$ be given. Let $d$ be the metric for $Y$.
Now as the open ball $$ B_d \left( \ f(x) , \frac{\varepsilon}{2} \right) \colon= \left\{ \ y \in Y \ \colon \ d \left( y, f(x) \right) < \frac{\varepsilon}{2} \ \right\} $$ is an open set in $Y$, as $f(x) \in B_d \left( \ f(x) , \frac{\varepsilon}{2} \right) $, and as the function $f \colon X \to Y$ is continuous, so the inverse image $$ f^{-1} \left( \ B_d \left( \ f(x) \ , \ \frac{\varepsilon}{2} \right) \ \right) \colon= \left\{ \ u \in X \ \colon \ f(u) \in B_d \left( \ f(x) \ , \ \frac{\varepsilon}{2} \right) \ \right\} = \left\{ \ u \in X \ \colon \ d \left( \ f(u) \ ,\ f(x) \ \right) < \frac{\varepsilon}{2} \ \right\} $$ is an open set in $X$ and also $x \in f^{-1} \left( \ B_d \left( \ f(x) \ , \ \frac{\varepsilon}{2} \right) \ \right)$.
Let $U \colon= f^{-1} \left( \ B_d \left( \ f(x) \ , \ \frac{\varepsilon}{2} \right) \ \right)$.
Then $U$ is an open set in $X$, $x \in U$, and also $$ f ( U ) = f \left( \ f^{-1} \left( \ B_d \left( \ f(x) \ , \ \frac{\varepsilon}{2} \right) \ \right) \ \right) \subset B_d \left( \ f(x) \ , \ \frac{\varepsilon}{2} \right), $$ by Prob. 1 (b) in the Exercises following Sec. 2 in Munkres; that is, $$ d \left( \ f(u), f(x) \ \right) < \frac{\varepsilon}{2} \tag{1} $$ for all points $u \in U$.
Now as $\left( x_n \right)_{n \in \mathbb{N} }$ is a sequence in the topological space $X$ converging to the point $x \in X$ and as $U$ is an open set in $X$ such that $x \in U$, so there is a natural number $M$ such that $x_n \in U$ for all natural numbers $n > M$. Therefore (1) implies that $$ d \left( \ f \left( x_n \right) \ , \ f(x) \ \right) < \frac{\varepsilon}{2} \tag{2} $$ for any natural number $n > M$.
Moreover, as the sequence $f_n$ of functions converges uniformly to $f$ on $X$, so there exists a natural number $N$ such that $$ d \left( f_n(t), f(t) \right) < \frac{\varepsilon}{2} $$ for all natural numbers $n > N$ and for all $t \in X$. Therefore $$ d \left( f_n(t), f(t) \right) < \frac{\varepsilon}{2} $$ for all natural numbers $n > N$ and for all $t \in U$. And, as $x_n \in U$ for all natural numbers $n > M$, so we can conclude that $$ d \left( \ f_n \left(x_n \right) \ , \ f \left( x_n \right) \ \right) < \frac{\varepsilon}{2} \tag{3} $$ for all natural numbers $n$ such that $n > M$ and $n > N$.
So for any natural number $n > \max \{ \ M \ , \ N \ \}$, we see that $$ \begin{align} d \left( \ f_n \left( x_n \right) \ , \ f(x) \ \right) &\leq d \left( \ f_n \left( x_n \right) \ , \ f \left( x_n \right) \ \right) + d \left( \ f \left( x_n \right) \ , \ f(x) \ \right) \\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\ & \qquad \mbox{ [ using (2) and (3) above; we note that } \\ & \ \ \ \qquad \mbox{ since $n > \max\{ \ M \ , \ N \ \}$, therefore $n > M$ and $n > N$ ] } \\ &= \varepsilon. \end{align} $$
Thus we have shown that, corresponding to every real number $\varepsilon > 0$, there exists a natural number $K \colon= \max \{ M , N \}$ such that $$ d \left( \ f_n \left( x_n \right) \ , \ f(x) \ \right) < \varepsilon $$ for any natural number $n > K$.
Hence the sequence $\left( \ f_n \left( x_n \right) \ \right)_{n \in \mathbb{N} } $ converges in the metric sapce $(Y, d)$ to the point $f(x) \in Y$, as required.
Is this proof correct? If so, then is each and every step in my presentation clear enough too? If not, then where lies the problem?
The whole essence of the proof is: for all $n$:
$$d(f_n(x_n), f(x)) \le d(f_n(x_n),f(x_n)) + d(f(x_n), f(x))$$
And both terms on the right can be made as small as we like: pick $n$ large enough for the uniform convergence definition for the first term, and $n$ also large enough from the fact that $f(x_n) \to f(x)$ by continuity of $f$.
A lot of your write-up is superfluous or repetitive, IMHO. Just convey the main idea.