= Solution
This is always true. Since $f$ is bounded and $f_n\to f$ uniformly, there are $M$ and $N$ such that
$$
|f(x)|\leq M,\qquad |f_n(x)|\leq M+1
$$
for every $x\in X$ and $n\geq N$. The continuous function $g$ is <uniform continuity>[uniformly continuous] on the compact interval $[-M-1,M+1]$. Therefore, given $\varepsilon>0$, a sufficiently small uniform bound on $|f_n-f|$ implies
$$
\sup_{x\in X}|g(f_n(x))-g(f(x))|<\varepsilon.
$$
\b[Hence $g\circ f_n\to g\circ f$ uniformly.]
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