I feel like I have written an "overkill" argument for this problem, but I wanted to be thorough.
Let $(X,d)$ and $(Y,\rho)$ be metric spaces, and let $f, f_{n} :X \to Y$ with $f_{n} ⇉ f$ on $X$ $($$f_{n}$ uniformly converges to $f$ on $X$$)$. Show that $D(f) \subset \bigcup_{n=1}^{\infty}D(f_{n})$, where $D(f)$ is the set of discontinuities of $f$.
Suppose that $x \in D(f)$. Then $\exists \epsilon > 0 \ \forall \delta >0$ such that $d(x_{n},x) < \delta$ and $\rho(f(x_{n}),f(x)) \geq \epsilon$. Now, since $f_{n} ⇉ f$ on $X$ we have that $\forall \epsilon > 0, \exists N \in \mathbb{N}$ such that $\rho(f_{n}(x),f(x)) < \epsilon, \forall x \in X, \forall n \geq N$. Therefore by the triangle inequality: $$ \rho(f(x),f(x_{n})) \leq \rho(f(x),f_{n}(x)) + \rho(f_{n}(x),f_{n}(x_{n})+\rho(f_{n}(x_{n}), f(x_{n}) < \epsilon+ \rho(f_{n}(x),f_{n}(x_{n}))+\epsilon$$ and so we have that $\rho(f_{n}(x),f_{n}(x_{n})) \geq \epsilon$. Therefore we have a sequence $x_n \to x$ such that $f_{n}(x_{n})$ does not converge to $f_{n}(x)$. So, $x$ is a point of discontinuity of $f_n$. So by definition $x \in \bigcup_{n=1}^{\infty}D(f_{n})$.
Is this a correct approach to this problem? Any criticism is welcome.