I'm trying to show whether the following statement is true or not:
If $X$ and $Z$ are matric spaces and $f_k:Z \to X$ is a sequence of continuous functions such that $f_k \to f$ pointwise (for some continuous function $f:Z \to X$) then the sequence $(f_k)$ converges uniformly to $f$ on $Z$.
I claim that the statement is true and here is my proof:
Let $z_0 \in Z$ and $\epsilon >0$ then there exist $\delta_1$, $\delta_2 >0$ and $k_0 \in \mathbb{N}$ such that:
\begin{equation} d_X(f_k(z),f_k(z_0)) < \frac{\epsilon}{3}, \text{ if } \ d_Z(z,z_0)< \delta_1 \\ d_X(f(z),f(z_0)) < \frac{\epsilon}{3}, \text{ if } \ d_Z(z,z_0)< \delta_2 \\ d_X(f_k(z_0), f(z_0)) < \frac{\epsilon}{3}, \text{ if } \ k \geq k_0 \end{equation}
So by taking $\delta = \max(\delta_1, \delta_2)$, $k \geq k_0$ and $d_Z(z,z_0)<\delta$ we get:
$$ d_Z(f_k(z),f(z)) \leq d_Z(f_k(z),f_k(z_0))+d_Z(f_k(z_0),f(z))+d_Z(f(z),f(z_0))<\epsilon$$
My question is the following: To prove uniform convergence the definition must be satisfied for all $z \in Z$, however my argument only considers those $z$ close to a fixed $z_0$. Could it be argued that since $z_0$ was arbitrary this works for all $z \in Z$? or How could I prove the uniform convergence?