= Solution
A <function> $f:(X,d_X)\to(Y,d_Y)$ is uniformly continuous if for every $\varepsilon>0$ there is $\delta>0$ such that
$$
d_X(x,x')<\delta
\implies d_Y(f(x),f(x'))<\varepsilon
$$
for all $x,x'\in X$.
Suppose $f_n\to f$ uniformly and every $f_n$ is uniformly continuous. Given $\varepsilon>0$, choose $N$ such that
$$
d_Y(f_N(x),f(x))<\varepsilon/3
$$
for every $x$. Uniform continuity of $f_N$ supplies $\delta>0$ such that $d_X(x,x')<\delta$ implies
$$
d_Y(f_N(x),f_N(x'))<\varepsilon/3.
$$
The triangle inequality then gives $d_Y(f(x),f(x'))<\varepsilon$. Thus the <uniform limit theorem for uniformly continuous functions> proves that $f$ is uniformly continuous.
<Pointwise convergence> is insufficient. On $[0,1]$, the uniformly <continuous functions> $f_n(x)=x^n$ converge pointwise to
$$
f(x)=\begin{cases}0,&0\leq x<1,\\1,&x=1,\end{cases}
$$
which is discontinuous and therefore not uniformly continuous.
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