I need help verifying my solution to a problem, it seemed really simple but that's often when I make mistakes so I'd appreciate some help. Let $D \subset \mathbb{C}$ and $f_n:D\rightarrow \mathbb{C},n\in\mathbb{N}$ be a sequence of continuous functions and let $f:D\rightarrow \mathbb{C}$ be a function. Let $z_0 \in D$ and $z_n\in D$ a sequence with $z_n \rightarrow z_0$ when $n\rightarrow \infty$. Show that $f_n(z_n) \rightarrow f(z_0)$ when $n \rightarrow \infty$ under the assumption that $f_n$ uniformly converges to $f$. My attempt:
Sequence of functions $f_n:D\rightarrow \mathbb{C}$ is continuous which means for all sequences $x_n \in D$ that converge to $x_0$, the following holds true $f_n(x_n) \rightarrow f_n(x_0).$ Since we assume that $f_n \rightarrow f$ converges uniformly we know that $f$ is continuous which means that by the definition of continuity $f(x_n) \rightarrow f(x_0)$ therefore it follows that $f_n(z_n) \rightarrow f(z_0).$
Since you seem to have the right ingredients, here is a hint to make your argument more precise: \begin{aligned}|f_n(z_n)-f(z)|&=|f_n(z_n)-f(z_n)+f(z_n)-f(z)| \\&\leq |f_n(z_n)-f(z_n)| + |f(z_n)-f(z)| \end{aligned}
Let $\epsilon \in (0, \infty)$. Since $f_n \to f$ uniformly, there is $N_1 \in \mathbb{N}$ so that the first term is $< \epsilon/2$ whenever $n \geq N_1$. By the Uniform Limit Theorem $\bf{*}$, there is $N_2 \in \mathbb{N}$ so that the second term is $<\epsilon/2$ whenever $n \geq N_2$. Taking $N=\max\{N_1, N_2\}$, we have that the sum is $<\epsilon$ whenever $n \geq N$.
$\bf{*}$: Be sure you are allowed to use the Uniform Limit Theorem here. If you are not then it wouldn't be too difficult to prove.