Is every complete metric space closed? I know that if $A\subset X$ where $X$ is a complede metric space, and $A$ is closed $\iff$ it's complete.
However is every metric space closed? E.g., can I take $X\subset X$ and since $X$ is complete, can I conclude it's closed?
 A: Being closed is relative to some space $Y$ that contains $X$. (By "contains" I mean not just the set containment but also that the metric on $X$ is the restriction of the metric on $Y$.) Technically, we should always say "closed in $Y$" instead of simply "closed", but often the ambient space $Y$ is clear from context so it is not mentioned. 
As Brian M. Scott said, every space $X$ is closed in $X$; this is a consequence of the definition of "closed" and has nothing to do with completeness. However, there is a connection with completeness:
A metric space is complete if and only if it is closed in every space containing it. 
Indeed, suppose $X\subset Y$  and $X$ is complete. Take any sequence in $X$ that has a limit in $Y$. Since it converges, it is a Cauchy sequence, hence it has a limit in $X$. This shows $X$ is closed in $Y$.
Conversely, suppose  $X$ is closed in any $Y$ that contains $X$. Let $Y$ be the completion of $X$. By construction, $Y$ is complete and $X$ is dense in $Y$. But since $X$ is closed in $Y$, it follows that $X=Y$, hence $X$ is complete.
