Uniformly convergent sequence proof. Prove that if $(f_k)$ is a uniformly convergent sequence of continuous real-valued functions on a compact domain $D\subseteq \mathbb{R}$, then there is some $M\geq 0$ such that $\left|f_k(x)\right|\leq M$ for every $x\in D$ and for every $k\in \mathbb{N}$.

My response: Basically, I am trying to show that uniform convergence on a compact domain implies uniform boundedness. Let $f(x)$ be the limiting function. Then I know that 
$\lim_{k\to\infty} \sup_{x \in D} | f_k (x) - f(x) | = 0$. Also, I know that $f$ is continuous, therefore it attains an absolute maximum $\in D$. How can I apply these two things to prove it?
You are very much on the right track. So I'll try to bump you the rest of the way there. 

We want to find a bound on all $f_k$ simultaneously. 

It is not too hard to simultaneously bound any finite collection of our functions. This means we can control any finite initial segment of our sequence fairly easily. 

As you have noted, $f$ is also continuous and hence has a maximum $M$. 

Then by uniform convergence have some $N$ large so that for $n\geq N$ we know that $$\lvert f(x) - f_n(x)\rvert &lt; 1$$ for all $x$. 

This gives us a bound on the tail of the sequence, they can be no bigger than $M + 1$. 

So now can you combine these controls over the tail and initial segment of our sequence to finish the proof?
I think this is quite obvious. Noted that $D$ is compact and $f$ is continuous so $f_k(D)$ is also compact for every $k$ and hence it is bounded for every $k\in\mathbb{N}$ and we can just take $M=\sup\bigcup f_k(D)$ where $M \ne +\infty$ as every $f_k(D)$ are bounded.