Given a completely metrizable space, say that it has property X if it can be embedded in some metric space such that its image is not closed. For example, the real line R can be embedded, topologically, in itself as (0,1) which is not closed. A compact space such as S^1, however, clearly cannot be embedded in any metric space in this way.

Is it true that a space has property X iff it can be embedded in itself in this way? My intuition says no, but I can't think of a counterexample offhand, partly because I don't know any interesting non-compact metric spaces.

Is property X equivalent to not being compact? If every non-compact (completely metrizable) space has a metrizable compactification then this is easy, but I don't know whether that is the case, although I suspect it might be. Which leads onto the following weaker question:

Given a space which does not have property X, can it be embedded in a space which does? Here my intuition says yes, this should be the case, but I can't think of a more general approach for constructing a suitable embedding space than compactification which, as I said, I can't prove will give a metrizable space. I don't think I know enough topology (I've only taken a couple of basic undergraduate courses).

Edit: 3. above should read "given a space which has property X, can it be embedded in a space which does not?