Is this claim harder to prove for arbitrary metric spaces than for the reals? Let $X$ and $Y$ be metric spaces. Define the distance between functions $f, g$ from $X$ to $Y$ as
$$d(f, g) = \sup_{x \in X} \frac{d(f(x), g(x))}{1+d(f(x), g(x))}$$
Is it true that if $f_n:X \to Y$ is uniformly continuous for each $n \in \mathbb{N}$, and $(f_n) \rightrightarrows f$ (i.e., converges uniformly) then $f$ is uniformly continuous?
I claim that it's true, and here's my proof. Let $\epsilon > 0$ be given. Since $(f_n) \rightrightarrows f$, there is some $N$ such that $d(f_N(x), f(x)) < \epsilon/3$ for all $x \in X$. Choose $\delta$ such that $d(f_N(x_1), f_N(x_2)) < \epsilon/3$ whenever $d(x_1, x_2) < \delta$. Then, whenever $d(x_1, x_2) < \delta$, we have $$d(f(x_1), f(x_2)) \leq d(f(x_1), f_N(x_1)) + d(f_N(x_1), f_N(x_2)) + d(f_N(x_2), f(x_2)) < \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon$$ Therefore $f$ is uniformly continuous.
This problem appears in Pugh's Real Mathematical Analysis. The problem has two parts. In part (a), the problem is posed for functions from $\mathbb{R}$ to $\mathbb{R}$, and then part (b) says, "What happens for functions from one metric space to another instead of $\mathbb{R}$ to $\mathbb{R}$?"
Now I'm confused that the problem has two parts and the second part does not appear to be different in any substantial way from the first part. This makes me think that my proof must be flawed, or that there is a really trivial proof for $X = Y = \mathbb{R}$ (i.e., easier than this one) which doesn't work for general metric spaces. Otherwise why would the question have two parts like that?

There is a step missing. You have to show that convergence in the function space metric $d$ (could you use a different symbol for that, $\tilde d$ or $d_F$ or ...) implies the usual uniform convergence for the function values.
That is, start with $d(f,f_N)<ϵ_0$, so that for any $x$ 
$$
1-\frac{1}{1+d(f(x),f_N(x))}<ϵ_0
$$ 
and conclude $$d(f(x),f_N(x))<\frac{ϵ_0}{1-ϵ_0}.$$
Then set $\frac{ϵ_0}{1-ϵ_0}=\fracϵ3$ and continue as you did.
