Let $\{f_n\}$ be a sequence of unif cont functions from $(X,d) \to (\Omega,\rho)$ s.t $\{f_n\} \to f$ uniformly. Show that $f$ is unif cont. 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.
 A: Your proof is mostly correct, but you don't need both $N_1$ and $N_2$.  Since the convergence is uniform, one $N$ works for all $x\in X$. And when you finally get to work and start writing $d(f(x),f(y))\leq\cdots$, you don't specify what $n$ is (does any $n\in\mathbb N$ work?).  Also you originally let $x,y\in X$ be arbitrary, but then specify that $d(x,y)<\delta_1$, which isn't good practice when writing up a formal proof. Here's how you could clean things up a bit:
Let $\varepsilon>0$ be given.  Since $f_n\to f$ uniformly, there is some $N\in\mathbb N$ such that $d(f_n(x),f(x))<\varepsilon/3$ for any $n\geq N$ and $x\in X$.  Fix $n\geq N$.  Since $f_n$ is uniformly continuous, there is some $\delta>0$ such that for any $x,y\in X$ with $d(x,y)<\delta$ we have $d(f_n(x),f_n(y))<\varepsilon/3$.  Now for any $x,y\in X$ with $d(x,y)<\delta$, we have 
\begin{align}
d(f(x),f(y))&\leq d(f(x),f_n(x))+d(f_n(x),f_n(y))+d(f_n(y),f(y))\\
&<\varepsilon/3+\varepsilon/3+\varepsilon/3=\varepsilon
\end{align}
and the result follows.
