Compact subset of a Banach space being Banach themselves

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So, let $X$ be a Banach space under norm $||\cdot||_X$ and $K\subset X$ a compact subset and a vector subspace of $X$. If we now restrict the norm from $X$ to $K$ (i.e. we take for $x \in K$, $||x||_K=||x||_X$ and by this define norm in $K$), will $K$ together with this norm be a Banach space?

What I think is that this should be positive, as $X$ is Banach space, it is before all, vector space, and $K$ , as a vector space, is vector space itself. Also, $||\cdot||_K$ is obviously a norm, and if sequence $\{x_n\}$ is Cauchy in K, it is as well Cauchy in $X$, because of how norms are defined. As $X$ is Banach, there is a $x = \lim_{n\to \infty} x_n$ and $x \in X$. But, there is a sequence in $K$ converging to that $x$ meaning $x$ is a point of accumulation of $K$. As $K$ is compact, it has to be closed, and hence contain all of its accumulation points, $x$ included. So, a random Cauchy sequence $x_n$ in $K$ converges to a point $x \in K$, and $K$ is Banach as well.

Am I right?? Because this is simple and important and still not in any textbook or wikipedia, and this shall be among the first things proven about Banach spaces??

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The only compact subspace is the trivial one $K = \{ 0 \}$. Yes, this is a Banach space, but not a particularly interesting one.

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Your (correct) proof shows: if $K$ is a compact subset of a Banach space, the $K$ is, with the metric $d(x,y)=||x-y||_X$, a complete metric space.

$K$ is not a Banach space, since $K$ is not a (vector) subspace of $X$