The answer is no.
Lemma: Every quasi-totally bounded (instead of finite, there is a countable net) metric space
is separable.
Proof: For every
let
be the guaranteed countable
-net and let
. Clearly
being the countable union of countable sets is countable. Also, we claim that it is dense in
. To see this let
and
be given. By the Archimedean principle there exist some
such that
. So, then since
cover
it must be that
for some
. Thus, for that
we have that
from where the conclusion follows.
It follows since every compact metric space is totally bounded that every compact metric space is separable.
Also:
Lemma: Let
be a
space with
dense. Then,
Proof: For a fixed
define
to be the set of all neighborhoods of
and define
. To see this is an injection we merely note that if
then
being
guarantees there are neighboorhoods
and
such that
and
. In particular since every neighborhood of
contains
we have that
and so
and so
. It follows that
is an injection from where the conclusion follows.
Thus, either looking
here, or noticing that
and so
and so by the above lemma
cannot be turned into a separable
space and thus not into a compact metric space.