From Rudin's Principles of Mathematical Analysis (p.37).
Theorem 2.33. Suppose $K\subset Y\subset X$. Then $K$ is compact relative to $X$ if and only if $K$ is compact relative to $Y$.
Immediately after the following is written.
By virtue of this theorem we are able, in many situation, to regard compact sets as metric spaces in their own right, without paying any attention to any embedding space. In particular, although it makes little sense to talk of open spaces, or of closed spaces (every metric space $X$ is an open subset of itself, and is a closed subset of itself), it does make sense to talk of compact metric space.
I can't understand the sentence "every metric space $X$ is an open subset of itself, and is a closed subset of itself".
Actually, $(0,1)$ is metric space for metric $d(x,y)=|x-y|$, $\forall x,y\in (0,1)$. $(0,1)$ is just open subset of itself, but is not a closed subset of itself.
To understand why this is true one must use the fundamental duality of openness and closeness. Note that if a set is open, it's complement is closed and vice versa. The complement of any metric subspace considered as a metric space in it's own right is the empty set. Clearly a metric space is closed with respect to itself since it contains all it's limits. This implies that the empty set is open in the metric space M. Also, the empty set is closed in M which implies M is an open subset of itself. Therefore every metric space is a clopen (open and closed subset of itself). Hope this helps.