It depends what you mean by "unit ball".
To be make the question precise, let
$$U:=\{x\in X: \|x\| < 1\}$$
and
$$E:=\{x\in X: \|x\| \leq 1\}.$$
Then the conclusions are the following;
(1) $U$ is open in $X$, and not closed in $X$.
(2) $E$ is closed in $X$ and not open in $X$.
(3) $E$ is compact if and only if $X$ is finite dimensional.
(4) $U$ is totally bounded in $X$ if and only if $X$ is finite dimensional.
(5) $E$ is complete in $X$ if and only if $X$ is complete.
The proofs of (1), (2) are easy, and is left to the reader as exercises.
For the proofs of (3) and (4) see the book: "Functional Analysis: A First Course", PHI-learning, New Delhi, 2002 (Fourth Print: 2014), by me (M.Thamban Nair).
Proof of (5) is also easy, recently observed by me, and is going to appear in an educational news letter. This shows that if the space is not complete, then its "closed unit ball" $E$ is not complete, and hence not compact.