If $K$ is a compact subset of $\mathbb{R}^d$, then $K$ contains a point of maximal norm. Just getting started with some basic topology so apologies if this seems elementary to all of the super smart math wizards about.
This makes sense to me intuitively but I am struggling to prove it. Here's what I have so far.

Theorem If $K$ is a compact subset of $\mathbb{R}^d$, then $K$ contains a point of maximal norm. That is, there is a point $a \in K$ such that $||x|| \leq ||a||$ for all $x \in K$.
Proof (so far) Since $K$ is compact, it is closed and bounded. Let $m = \sup \{||x|| : x \in K\}$. Then the set of open balls $B_{m - 1/n}(0)$ is an open cover of $K$.
Since $K$ is compact, there is a finite subcover of $K$ from the set of open balls $B_{m - 1/n}(0)$, i.e., there exists an $N$ such that when $n \geq N$,$B_{m - 1/n}(0)$ covers $K$.

My next thought was to then consider $ K_0 = B_{m - \frac{1}{N}}(0) - B_{m-\frac{1}{N-1}}(0)$. If it contains exactly one point of $K$, then that is the point of maximal norm $a \in K$ we're looking for.
Where I'm stuck is if $K_0$ contains more than one point of $K$. I figure I could let $m_0 = \sup \{||x|| : x \in K_0 \cap K\}$, which leads to the existence of an $N_0$ such that when $n \geq N_0$, $B_{m_0 - \frac{1}{n}}(0)$ covers $K_0$, but then I'm right back to the same issue of whether $K_0$ contains one or multiple points of $K$, so I don't think that's the right approach. It also just occurred to me that $m$ and $m_0$ are the same.
I'm sure I'm making this much more difficult than it needs to be. Any suggestions?
 A: By your definition of $m$ , let $\{x_n \} \subset K $ s.t $||x_n|| \to m$. 
There is such $x_n$ s by the definition of supremum.
$K$ is compact so you can find $x_{n_k} \to x\in K$. 
Not try to show that $||x|| = m$ and you are done.
A: Your idea works, up to some minor modifications. First, in order to get that $U_n := B_{m - \frac1n}(0)$ is an open cover for $K$, you have to assume for a contradiction that no such $a$ exists. Let's do that. Then for each $x \in K$, $\|x\| < m$ so that there exists an $n$ such that $x \in U_n$ so that the $U_n$ are an open cover.
Take a finite subcover $C = \{U_{n_1}, \dots, U_{n_k}\}$ . Suppose without loss of generality that $n_1 < \dots < n_k$ so that $K \subseteq U_{n_k} = B_{m-\frac{1}{n_k}}(0)$. But then for $x \in K$, $\|x\| < m - \frac{1}{n_k}$ which contradicts the definition of $m$.
A: Try arguing by contradiction by supposing there is no such a and see if in that case there is still a finite subcover of that specific open cover.
That is an easier way to show this result.
A: Your proof as commented on in the other answer, will work. Here is another way to do it: let $f:=\|\cdot \|:\mathbb R^d\to \mathbb R_{\ge 0}$ be a norm on $\mathbb R^d.$ An application of the triangle inequality shows that $f$ is continuous. Suppose that $f$ has no maximum on $K$. Then, $\{(-\infty,f(x))\}_{x\in K}$ is an open cover of $f(K)$. Since $f$ is continuous, $f(K)$ is  compact, so there are integers $i=1,\cdots, n$ such that $\{(-\infty,f(x_i))\}^n_{i=1}$ covers $f(K).$ Without loss of generality, label the $f(x_i)$ so that they are in increasing order. Then, $f(x_n)\notin f(K),$ which is a contradiction.
