1. ## Uniform Convergence

I would appreciate any help that I can get with the following proof:

Let $f_n,f:[0,1]\rightarrow \mathbb{R}$ and $x,x_n\in [0,1]$ such that each $f_n$ is continuous. Also, $f_n\rightrightarrows f$ and $x_n\rightarrow x$.
Prove that $f_n(x_n)\rightarrow f(x)$.

So i think my goal is to find an $N$ such that $n\geq N\Rightarrow |f_n(x_n)-f(x)|<\epsilon$.
Also, we know that there exists an $N$ such that $n\geq N\Rightarrow |f_n-f|<\epsilon$ and also, $|x_n-x|<\epsilon$.
Assuming I'm correct so far, I need some help putting it all together...

2. $|f_n(x_n)-f(x)|\leq |f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|$.

3. So could we say that since $f_n\rightrightarrows f$ then $|f_n(x)-f(x)|<\epsilon$ for any $x\in [0,1]$. Thus, since $x_n\in [0,1]$ we have that $|f_n(x_n)-f(x_n)|<\epsilon$ must be true. Then, since we know $f$ is continuous and $x_n\rightarrow x$ we have that $|f(x_n)-f(x)|<\epsilon$ must also be true.
Then, we have that $|f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|<2\epsilon$. Then, by the triangle inequality, $|f_n(x_n)-f(x)|\leq|f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|<2\epsilon$. Thus, we have that $f_n(x_n)\rightarrow f(x)$.