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The notes for my functional analysis class casually state that every nontrivial normed space is not compact, but do not give a proof. I can't tell if this is because it is a trivial proof or because it is too complicated to be given inline. Is there a good proof for this? I have searched around the internet for a while and haven't been able to find one.

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    Consider the open cover of unit balls centered at every point in the space. – 2017-02-10
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    The result is false. Every zero dimensional normed space is compact. – 2017-02-10
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    All $1$ of them :P – 2017-02-10
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    If we're going to be hyperpedantic, there's only one up to canonical isomorphism. – 2017-02-10

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Any compact subset of any metric space is bounded.

Since any nonzero normed space is unbounded, it isn't compact.

EDIT

Consider any non zero Vector $x$ and any $M>0$; we have $\Vert tx\Vert=\vert t\vert\Vert x\Vert$ which Will be $>M$ for suitable $t$

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    It would be helpful if you demonstrated that it is indeed unbounded. – 2017-02-10
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    @AndresMejia: Ok ... Editing – 2017-02-10
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Of course, the trivial normed space $\{0\}$ with norm $\|0\|=0$ is compact. So let's restrict our attention to normed spaces with dimension $\geq1$.

Let $X$ be a nontrivial normed space, and let $x\in X$ be nonzero. Now consider the sequence $\{nx:n\in\mathbb N\}$. For each $n$ the set $$B_n=\{y\in X:\|nx-y\|<\frac{1}{2\|x\|}\} $$ is an open neighborhood of $nx$ which does not contain any other point in the sequence. Thus no subsequence is convergent (why?). Hence $X$ is not sequentially compact, and therefore $X$ is not compact.