We know that for a space $X$ if $X$ is reflexive then $X^*$ (the dual space) is reflexive. We also know that if $X$ is a banach space then if $X^*$ is reflexive then $X$ is reflexive.
Is the requirement for $X$ to be banach necessary?
Can we find a non-reflexive space $Y$ such that $Y^*$ is reflexive?
To expand on the idea hinted at in the comments, let $Y$ be a dense proper subspace of a reflexive Banach space $X$. $Y$ is not reflexive, because $Y$ is not complete and any normed space which is reflexive is a Banach space. However, it can be shown that $Y^{*}$ is isomorphic to $X^{*}$. This is because each element of $Y^{*}$ can be extended uniquely to obtain an element of $X^{*}$ with the same norm, while the restriction of each element of $X^{*}$ to the subspace $Y$ obtains an element of $Y^{*}$ with the same norm. Because $X$ is reflexive, $X^{*}$ is also reflexive. So $Y^{*}$ is reflexive, because the property of being reflexive is preserved under isomorphisms.
For a concrete example, consider $\ell_{2}$ equipped with the usual $\ell_{2}$ norm, and $c_{00}$, the vector space of sequences with finitely many non-zero terms, equipped with the induced norm from $\ell_{2}$. $c_{00}$ is a proper dense subspace of $\ell_{2}$ and is not reflexive itself. But because the dual of $\ell_{2}$ is isomorphic to itself, the dual of $c_{00}$ with its induced norm is isomorphic to $\ell_{2}$, so the dual of $c_{00}$ equipped with this norm is reflexive.
To see why each element of $Y^{*}$ can be uniquely extended to obtain an element of $X^{*}$, take a look at this post.
