$\phi : \mathbb{R}\rightarrow \mathbb{R}$ is a function with the property that whenever a sequence $f_n : \mathbb{R}\rightarrow\mathbb{R}, n=1,2,...$ converges uniformly, so does $\phi\circ f_n$. Show $\phi$ is uniformly continuous.
Attempt) Suppose $\phi$ is not uniformly continuous. Then there is an $\varepsilon>0$ such that for every $\delta>0$, there exist $x,y\in\mathbb{R}$ with $\mid x-y\mid<\delta$ but $\mid\phi(x)-\phi(y)\mid\geq\varepsilon$. So for $n=1,2,...$, there exist $x_n,y_n$ such that $\mid x_n-y_n\mid<\frac{1}{n}$ but $\mid\phi(x_n)-\phi(y_n)\mid\geq\varepsilon$. I am trying to find a uniformly convergent sequence $f_n$ for which $\phi\circ f_n$ does not converge uniformly to draw a contradiction.