Any metric space can be isometrically embedded into some Banach space *E*. And *E* has the property X, since it can be topologically embedded into itself as its open unit ball, which is clearly not closed in *E*. Therefore, the answer to *3.* is **YES**, and the intuition of Robin Saunders is very good.
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This is just to clarify some things. For this, let $(M,d)$ be a complete
metric space.

It is pretty clear that a compact metric space cannot have "the
property X". Conversely, asumming that $M$ is non-compact and that $d$ is bounded, by taking
into consideration the equivalent metric $d^{*}\left(x,y\right)$= $\sup_{t\in M}$ $\mid f(x)\cdot d(x,t)$ $-f\left(y\right)\cdot d\left(y,t\right)\mid$, where $f:M$ $\rightarrow$ $(0,\infty)$
is bounded and continuous and inf$_{M}$ $f$ = 0, one easily obtain that $(M,d^*)$
is not complete.

Consequently, $M$ has "the property X" iff it is non-compact.

Now, every separable metric space can be embedded into the Hilbert
cube, which has not "the property X" .

[Note that on [Wikipedia]
you'll find "every separable metric space can be isometrically
embedded into the Hilbert cube", which is obviously not true.]

OTOH, any subspace of a separable metric space is separable, too.
And any compact metric space is, clearly, separable.

Therefore, $M$ is embeddable into a space having not "the property
X" iff $M$ is separable.


  [Wikipedia]: http://en.wikipedia.org/wiki/Hilbert_cube