Sum of compact sets

Remember $+$ is a function defined as

$$+:\begin{cases} A\times B\to A+B \\ (a,b)\mapsto a+b\end{cases}$$

Since addition is continuous on normed spaces and $A+B$ is the image of the compact set $A\times B$ under this map, the image is compact.

Adam's proof is the cleanest. Here is an alternative proof:

Suppose $x_n \in A+B$ is Cauchy. We have $x_n = a_n+b_n$, with $a_n \in A$ and $b_n \in B$. Since $A,B$ are compact, we have $a_{n_k} \to a \in A$ and $b_{n_k} \to b \in B$ for some subsequence $n_k$. Hence $x_{n_k}=a_{n_k}+b_{n_k} \to a+b \in A+B$, and since $x_n$ is Cauchy, we have $x_n \to a+b$. Hence $A+B$ is complete.

Let $\epsilon>0$ and choose finite ${ 1 \over 2}\epsilon$-nets for $A,B$. That is, some finite collection $a_k,b_k$ such that for any $a \in A,b\in B$ there is some $k,k'$ such that $\|a-a_k\| < { 1 \over 2}\epsilon$, $\|b-b_{k'}\| < { 1 \over 2}\epsilon$. Let $x \in A+B$, then $x=a+b$ for some $a \in A,b\in B$. As above, there is some $k,k'$ such that $\|x-(a_k+b_{k'})\| = \|a+b-(a_k+b_{k'})\| \le \|a-a_k\|+ \|b-b_{k'}\| < \epsilon$. Hence the collection $a_k+b_{k'}$ form a finite $\epsilon$-net for $A+B$. Hence $A+B$ is totally bounded.