Let $V$ be a vector space, and assume that $V$ is isomorphic to its dual, i.e., $V \simeq V^*$. Is every linear subspace $U$ of $V$ also isomorphic to its dual, i.e., $U \simeq U^*$?
This is certainly true in finite dimensions, and I believe also for Hilbert spaces, so assume $V$ is infinite-dimensional and not a Hilbert space. I believe that, if there is any chance of the above being true, we also need to assume that $V$ is a topological vector space, $V^*$ is the continuous/topological dual space (rather than algebraic dual space), and the isomorphism $V \simeq V^*$ is continuous linear (with continuous linear inverse).
I assume that you're referring to the continuous dual of a normed (or at least topological) vector-space.
If we don't specify that $U$ is a closed subspace, then the answer is no, not even for Hilbert spaces. As an example, take $V = \ell^2$ and $U = c_{00} \subset V$. $c_{00}$ is separable, but $c_{00}^* \cong \ell_\infty$ is not.