Consider the sequence of functions $f_k :A \rightarrow \mathbb{R}$. Suppose that $\{f_k\} \rightarrow f$ uniformly in A and $f_k$ is continuous in A $\forall k$. Prove that if $x_0 \in A$ and $\{x_k \}$ is a sequence of points of $A$ such that $\{x_k \} \rightarrow x_0$, then $\{f_k(x_k) \} \rightarrow f(x_0)$.
This is what i' ve done so far:
I have already proven that if a sequence $f_k$ converges uniformly to a function $f$ and $f_k$ are continuous $\forall k$, then $f$ is continuous.
Since $f_k \rightarrow f$ uniformly, given $\epsilon >0$, there exists $n_0 \in \mathbb{N}$ such that $\forall n \geq n_0$, $|f_n(x)-f(x)|<\epsilon$ for any x, so if we consider the sequence $\{x_k \} \rightarrow x_0$, $|f_n(x_i)-f(x_i)|<\epsilon$ for all $x_i \in \{x_k \}$ so $\lim_{i\to \infty} |f_n(x_i)-f(x_i)|=|f_n(x_0)-f(x_0)|<\epsilon$. But this would only prove that $\{f_k(x_0) \} \rightarrow f(x_0)$, which is not what I' m trying to prove...
Hint : Write
$|f_k(x_k)-f(x_0)|=|f_k(x_k)-f(x_k)+f(x_k)-f(x_0)| \leq |f_k(x_k)-f(x_k)|+|f(x_k)-f(x_0)|$
$\leq \sup_{x \in A}|f_k(x)-f(x)|+ |f(x_k)-f(x_0)|.$