Uniform continuous maps that converge uniformly to a function f. What can we say about f? Here is the question:
Let $X$ and $Y$ be metric spaces.Suppose a sequence of uniformly continuous maps $f_n : X \rightarrow Y$ converges uniformly to a map $f: X \rightarrow Y$.Does that imply that f is continuous ? Uniformly continuous ($X$ is not necessarily compact)?
I proved (hopefully right) that $f:X \rightarrow Y$ is continuous , even without using uniform continuity of $f_n$ but just assuming $f_n$ are continuous. 
I can't prove or disprove that $f$ is uniformly continuous  and whether adding the compactness condition would change the answer.

Given an $\epsilon > 0$, first find an $N$ such that $d_Y(f(x),f_N(x))< \frac{\epsilon}{4}$ for all $x\in X$. This is possible because of uniform convergence.
Next, find a $\delta > 0$ so that for any $x, y \in X$ with $d_X(x, y) < \delta$ gives $d_Y(f_N(x), f_N(y)) < \frac{\epsilon}{2}$. This is possible by uniform continuity of $f_N$. We now have, for the same $x, y$ that
$$
d_Y(f(x), f(y)) \leq d_Y(f(x), f_N(x)) + d_Y(f_N(x), f_N(y)) + d_Y(f_N(y), f(y)) \\\\
 < \frac{\epsilon}{4} + \frac{\epsilon}{2} + \frac{\epsilon}{4} = \epsilon
$$
thus we have found a $\delta$ so that for any $x, y \in X$ with $d_X(x, y)<\delta$ we have $d_Y(f(x), f(y)) < \epsilon$. Our $\epsilon$ was arbitrary, and therefore $f$ must be uniformly continuous.
