Always Closed Metric Space is Complete

I am trying to prove that

Given a metric space $Y$, if for every metric space $X \supseteq Y$, $Y$ is $X$-closed, then $Y$ is complete.

by contradiction or contraposition, such that I don't use the existence of a completion. (This is 3.3:1 (c) in Bergman's supplementary exercises to Rudin PMA).

Supposing that $Y$ is not complete, I think that the divergent Cauchy sequence in $Y$ should intuitively tend to some $p \not \in Y$. The difficulty, however, is that the ambient space could not be specified until this $p$ is found.

Or could looking at its set of subsequential limits help?