This is what I have so far:
$\textbf{Proof:}$
We must show that $f$ is uniformly continuous, hence satisfy that
$\forall \epsilon > 0, \exists \delta > 0, \forall x,y \in X$ with $d(x,y) < \delta \Rightarrow \rho(f(x),f(y)) < \epsilon$.
Let $\epsilon > 0$ and $x,y \in X$. Since each $f_n$ is uniformly continuous we have that
$\exists \delta_1$ such that $\rho(f_n(x),f_n(y)) < \frac{\epsilon}{3}$, whenever $d(x,y) < \delta_1$.
We also have that $f_n \to f$ uniformly, then
$\exists N_1 \in \mathbb{N}, \forall x \in X$, such that $\rho(f_n(x),f(x)) < \frac{\epsilon}{3}$ whenever $n\geq N_1$.
$\exists N_2 \in \mathbb{N}, \forall y \in X$, such that $\rho(f_n(y),f(y)) < \frac{\epsilon}{3}$ whenever $n\geq N_2$.
By the triangle inequality
$\rho(f(x),f(y)) \leq \rho(f(x),f_n(x)) + \rho(f_n(x),f(y))$
Applying again the triangle inequality yields
$\rho(f(x),f(y)) \leq \rho(f(x),f_n(x)) + \rho(f_n(x),f_n(y)) + \rho(f_n(y),f(y))$
Hence,
$\rho(f(x),f(y)) \leq \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$.
I know there is something weird (incomplete) since I didn't use the $\delta_1$ and also I didn't specify how to choose the $\delta$. My problem is how to relate the $\delta$ with the $\delta_1, N_1$ and $N_2$.
Any help would be appreciated. Thanks.