Let $f:\mathbb R \to \mathbb R$ be a continuous function. Define $f_n:\mathbb R \to \mathbb R$ by $$ f_n(x) := f(x+1/n). $$ Suppose that $(f_n)_{n=1}^\infty$ converges uniformly to $f$. Does it follow that $f$ is uniformly continuous?
Note: the answer is clearly no if we don't assume that $f$ is continuous. I suspect there is a counterexample, showing that the answer is no even if $f$ is continuous.
Edit: The following observation might help.
There exists a continuous, non-uniformly continuous function $f:\mathbb R \to \mathbb R$ such that $(f_n)_{n=1}^\infty$ converges uniformly to $f$, if and only if there exists a continuous, non-uniformly continuous function $g:\mathbb N \times \mathbb R \to \mathbb R$, such that $(g_n)_{n=1}^\infty$ converges uniformly to $g$, where $g_n(k,x) = g(k,x+1/n)$ and the metric on $\mathbb N \times \mathbb R$ comes from viewing it as a subspace of $\mathbb R \times \mathbb R$ (with the Euclidean metric, say).
Proof: Given the function $g$, there is some $\epsilon>0$ which witnesses non-uniform continuity. That is, for each $n$, there exists $m \in \mathbb N$ and $x,y \in \mathbb R$ such that $|x-y|<1/n$ and $$|g(m,x)-g(m,y)|\geq\epsilon.$$ By moving things around, we can assume that for each $k$, there exists $y \in (0,1/k)$ such that $$ |g(k,0)-g(k,y)| \geq \epsilon. $$
Next, we can also assume that $g(k,x)=0$ for all $|x|>2$. This is achieved by multiplying $g$ by the function $h$ which is $1$ on $\mathbb N \times [-1,1]$, $0$ on $\mathbb N \times (\mathbb R \setminus (-2,2)$, and linear elsewhere. That $(gh)_n \to gh$ uniformly can be proven by the same argument that the product of uniformly continuous functions is uniformly continuous.
Now, we just define $f$ piecewise, by [ f(6k+x)=g(k,x) ] for $k \in \mathbb N$ and $x \in [-3,3]$, and [ f(x) = 0 ] for $x < -3$.