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 $(X',d')$.