$f_n(x_n) \rightarrow f(x) $ by uniform convergence I am very nearly done with this problem, but I have a concern that someone must help me alleviate. 

Suppose $f_n\rightarrow f$ uniformly, $f_n$ are continuous,  and $x_n\rightarrow x$. Prove that $$\lim_{n\to \infty} f_n(x_n) =f(x) .$$

We want to use $$|f_n(x_n) - f(x) |\leq |f_n(x_n) - f_n(x) |+|f_n(x) - f(x) |.$$ 
At first, I let $N$ be a number so that $n\geq N$ implied that $|f_n(x) - f(x) |< \epsilon /2$ for any $x$. Now,  since $f_n$ is continuous, $\lim_{m\to\infty} f_n(x_m) =f_n(x) $, we can take $|f_n(x_m) - f_n(x) |<\epsilon /2$ for $m$ greater than some $M$. But, can we take $|f_n(x_n) - f_n(x) |<\epsilon /2$? Maybe not,  if $n<M$. We could insist this is not true, and take $n$ to be greater than $M$ by WLOG letting $N$ get larger. But since we defined $M$ in the context of a particular $f_n$, now $M$ might be different,  so we are going in a circle. How can I fix this? 

I will give a proof using a slightly different version of the triangle inequality than the one used by OP namely
$$|f_n(x_n) - f(x)| \leq |f_n(x_n) - f(x_n)| + |f(x_n) - f(x)|$$
We start off by fixing $\epsilon > 0$. From the definition of uniform convergence there exist a $N$ s.t. if $n>N$ the first term is less than $\epsilon / 2$.
The uniform convergence of $f_n\to f$ and the fact that $f_n$ is continuous it follows that $f$ is continuous. This implies that there exist a $\delta>0$ s.t. if $|x-x_n|<\delta$ then $|f(x_n) - f(x)| < \epsilon/2$. Finally since $x_n\to x$ there exist a $M$ s.t. if $n>M$ we have $|x-x_n| < \delta$.
Putting it all togeather we find that if $n>\text{max}(N,M)$ then
$$|f_n(x_n) - f(x)| < \epsilon/2 + \epsilon/2 = \epsilon$$
