Question about applying a sequence of continuous functions that converge uniformly to a convergent sequence of complex numbers Let $(w_n)$ be a sequence of complex numbers in the closed unit disk
$$D = \{ z \in \mathbb{C} \ \ | \ \ |z| \leq 1 \}$$
that converges to $w$. Suppose that $(f_n)$ is a sequence of continuous functions from $D$ to $\mathbb{C}$ that converges uniformly on $D$ to a function $f$. Prove or disprove that 
$$\lim_n f_n(w_n) = f(w).$$
My working so far: On the expectation that the statement is true, I have tried attacking it directly using the usual $\epsilon$'s and $N$'s, but to no avail. I have also tried to show that $(f_n(z_n))$ is a Cauchy sequence but that does not seem to work either.
 A: Hint: Using the $\varepsilon$-$\delta$ definition is a good way to go. Here's an inequality that'll help you:$$|f_n(w_n)-f(w)|\le|f_n(w_n)-f(w_n)|+|f(w_n)-f(w)|.$$
A: Let us fix $\epsilon > 0 $.
We want to prove, that there exists $ N $, such that for all $n>N$ we have $|f_n(\omega_n) - f(\omega)| < \epsilon $
You know, that since $f_n$ converges to $f$ on D, and $\omega \in D$, there exists $N_f$, such that for all $n>N_f$, we have $|f_n(\omega) - f(\omega)| < \frac{\epsilon}{2} $ $ \ \ \ \ \ (*)$
Similarly, we want to show $|f_n(\omega_n) - f_n(\omega)| < \frac{\epsilon}{2}$ for $n>N_\omega$. Why does there exist such a $N_\omega$ ?. Because, we know that $\forall_n$ $f_n$ is a continuous function on D and $\omega_n$ converges to $\omega$ (so, there exists $N_\omega$ that for every $n>N_\omega$ we have $|\omega_n - \omega| < \delta_\omega  \ \ $ (for particular $\delta_\omega$), and now by continuity of $f_n$, since $|\omega_n - \omega| < \delta_\omega$, we get $|f_n(\omega_n) - f_n(\omega)| < \frac{\epsilon}{2}$ $ \ \ \ \ \ \ \ (**) $
Now, taking $ N = max\{N_f,N_\omega\} $, we get that $\forall_{n>N}$ $$ |f_n(\omega_n) - f(\omega)| = |f_n(\omega_n) - f_n(\omega) + f_n(\omega) - f(\omega)| <^{triangle}_{inequality}  \\ < |f_n(\omega_n) - f_n(\omega)| + |f_n(\omega) - f(\omega)| <^{(*)}_{(**)} \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon  $$
Q.E.D
