Let $f_n, f,g \colon \mathbb R \to \mathbb R$ be continuous. Assume that $f_n \to f$ uniformly. Does it follow that $g(f_n) \to g(f)$ uniformly? Let $f_n, f,g \colon \mathbb R \to \mathbb R$ be continuous. Assume that $f_n \to f$ uniformly. Does it follow that $g(f_n) \to g(f)$ uniformly? Here $g(f)$ stands for the composition of $g$ and $f.$
I would expect the answer to be negative as in order to get the uniform convergence of $g(f_n) \to g(f)$ I would expect to require $g$ to be uniformly continuous.
 A: I would suggest a different solution than that of @Sean song since I think there is an easier counterexample.
Take $g(x)=e^x$, $f_n(x)= x+\frac{1}{n}$ and $f(x)=x$. Then $\Vert f_n-f\Vert_\infty=\frac{1}{n}\to 0$.
Since $g'(x)=g(x)$, we know by the mean value theorem that for any $a<b$, there exists a $c\in (a,b)$ satisfying
$$ e^c(b-a)= e^a(1-e^{b-a}) . $$
Hence, for any sequence $\{ x_n \}$ we have
$$ \Big\vert e^{c_n}\cdot \frac{1}{n} \Big\vert= \big\vert e^{x_n} \cdot(1-e^{\frac{1}{n}}) \Big\vert=\vert g\circ f(x_n)-g\circ f(x_n)\vert$$
for some $x<c_n<x+\frac{1}{n}$. So long as $\frac{e^{c_n}}{n} \geq \frac{e^{x_n}}{n} \geq  1$ for all $n\in \mathbb{N}$, this will show that there is no uniform convergence. Choosing $x_n=n$, will do the trick since
$$ \frac{e^x}{x}\to \infty \quad \text{as } \; x\to \infty. $$
A: EDIT : As pointed out in @keen-ameteur and my own comments below, there is a mistake in the answer - to find a counter example I cannot use $g$ with domain $x\geq2,$ for then since $f_n\to0$ we cannot assess at all how $g(f_n)$ behaves. Thank you very much for pointing this out.

Original answer :
Consider the functions $f_n(x) = 1/x^n$ defined on $(2,\infty)$. Now these functions converge uniformly to $f(x) = 0$. To see this, let $\epsilon$ be given. Note that
$$\|f_n(x) - 0\| = \frac{1}{x^n} < \frac{1}{2^n},$$
And thus there exists $N$, independent of $x$, such that $n>N$ implies $\|f_n(x)-0\|\leq \epsilon$ for all $x\in (2,\infty).$
However, if we define $g(x) = 1/x\,(x\geq2),$, then we have $g(f_n(x)) = x^n$. It follows that your sequence $g(f_n)$ does not even converge.
EDIT: To extend the domain to $\mathbb{R}$, I suppose we may simply extend the functions to be symmetric about $x=2$.
I guess i'll elaborate on my answer. Let
$$g(x) = \begin{cases} 1/x & x\geq 2 \\ 1/(4-x) & x<2.\end{cases}$$
Then $g$ is continuous everywhere on $\mathbb{R},$ and the argument does not break. It is easy to extend the $f_n$ in a similar fashion, define $$f_n(x) = f_n(4-x)$$ whenever $x\leq 2.$
Since the composition of symmetric functions is again symmetric, it follows that $g\circ f_n$ turns out to be the graph of
$$y=x^n,\, x\geq 2$$
flipped around the axis $x=2.$ As $n\to \infty$, we have $g(f_n(x)) \to \infty$ for all $x\in \mathbb{R}$.
