 Solution

This is always true. Since $f$ is bounded and $f_n\to f$ uniformly, there are $M$ and $N$ such thatfor every $x\in X$ and $n\ge N$. The continuous function $g$ is uniformly continuous on the compact interval $[-M-1,M+1]$. Therefore, given $\epsilon >0$, a sufficiently small uniform bound on $|f_n-f|$ impliesHence $g\circ f_n\to g\circ f$ uniformly.