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Let $(X,d)$ be a metric space. Assume that $(X', d')$ is a metric space such that $X\subseteq X'$ and $d'|_X=d$. (Basically, there is an embedding $(X,d)\hookrightarrow (X',d').$) Let $K\subseteq X$ be compact.

Is it true that $K$ is compact in $X'$ ?

My idea:

In a metric space compactness is equivalent to sequential compactness. Since $K$ is sequentially compact in $(X,d)$, when $K$ is viewed as a subset of $(X,d)$ every sequence in $K$ has a convergent subsequence which converges inside $K$. When $K$ is viewed as a subset of $(X',d')$, the same subsequence will work. Hence $K$ is compact in $(X',d')$.

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    $\begingroup$ Looks good to me. Notice compactness is a property of the topology on $K$. The surrounding space is rather unimportant. $\endgroup$ – user251257 Nov 24 '16 at 14:43
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    $\begingroup$ I agree with @user251257 (especially his or her emphasis on compactness being an intrinsic property): you have answered your own question. Incidentally, it's almost as easy to argue directly via the definition of compactness, without any appeal to metric properties. $\endgroup$ – LSpice Nov 24 '16 at 19:03

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