Uniform Convergence on Interior implies Uniform Convergence Everywhere First, a motivating toy case. Suppose that $\{f_n(x)\}$ is a pointwise-convergent sequence of functions on $[0,1]$ which converges uniformly on $(0,1)$. Then in fact the sequence converges uniformly on the entire interval. 
The proof of this is simple. Since there is uniform convergence on the interior, there is some $N$ for which the $f_n$ are within $\epsilon$ on the interior. But since there is pointwise convergence on the boundary, we can find $N_1$ and $N_2$ so that these are also within $\epsilon$. Thus we may take a new $N$ which is just the maximum of these three.
Now in general, this proof won't work. Replacing the interval by some compact set $K$ in a metric space $X$, the interior of $K$ may have infinitely many limit points in its boundary, each with different $N$, so a priori there is no reason to expect that the same kind of argument will work, but I have been unable to find a different one that will.
Because this argument is fundamentally broken as far as I can see, I am skeptical if the claim is even true in general, but I also couldn't find a counter-example.
 A: Simple counterexample: Let $A=\{1/k: k=1,2,\dots \}.$ Define $K=[-1,0] \cup A.$ Set $f_n(x) = 0,$ $x\in [-1,0],$ $f_n(x) = x^n,$ $x\in A.$ Then the $f_n$ are continuous on $K$ and the $f_n$ converge pointwise on $K$ to $\chi_{\{1\}}.$ The $f_n$ converge uniformly on the interior of $K,$ which is $(-1,0),$ but not on the boundary, which contains $A.$
On the other hand, if $K$ is the closure of its interior, and if the $f_n$ are continuous on $K,$ then uniform convergence on the interior implies uniform convergence everywhere. That's simply because every point on the boundary is the limit of a sequence from the interior. So  if $|f_n-f_m|<\epsilon$ on the interior, you'll have $|f_n-f_m|\le\epsilon$ on $K$ by taking limits.
A: Assuming that the $f_n$ do not need to be continuous on $K$, we have a simple counterexample. Consider any $\{g_n\}_{n=1}^{\infty}$, $g_n : S_1\to \mathbb{R}$, such that $g_n\to g$ pointwise but not uniformly. Let $K\subset \mathbb{R}^2$ such that $\partial K$ is a Jordan curve (and therefore we have a homeomorphism $\varphi : S_1\to \partial K$). Then, let $f_n$ be defined such that $f_n$ converges uniformly in $\operatorname{Int}(K)$ and $f_n = g_n\circ \varphi^{-1}$ on $\partial K$. Then, $f_n$ converges pointwise but not uniformly to $g\circ \varphi^{-1}$ on $\partial K$.
As pointed out by @zhw, if the $f_n$ are continuous on $K = \overline{\operatorname{Int}(K)}$, then uniform convergence on all of $K$ is immediate by taking the limit of $\lvert f(x)-f_n(x)\rvert$ as $x$ approaches $\partial K$.
