Is $X$ reflexive if $X=\cup Y_i $ and $Y_i$ is reflexive for every $i$? Let $X$ be a Banach space. Assume that there are a family of closed (with respect to $\left\|\cdot\right\|_X$) subspaces $Y_i$ of $X$ such $X=\cup Y_i $ and $Y_i$ is reflexive for every $i$. Can we say that $X$ is reflexive? (otherwise, under what condition,  $X$ is reflexive).
The answer is (somewhat surprisingly) yes. The reason it's yes is because the assumption $X=\bigcup\limits_{i}Y_i$ is a very strong assumption which greatly reduces which sets $Y_i$ we can consider. Assume that $X = Y_1\cup Y_2$ as a simple example. Then, without loss of generality, $Y_2\subset Y_1$ since otherwise we can take $y_2\in Y_2\backslash Y_1$ and $y_1\in Y_1\backslash Y_2$ to have that $y_1+y_2\in X$ but $y_1+y_2\not\in Y_1\cup Y_2$.

So if $X$ can be written in this way, there must be a "biggest" $Y_k$ which contains all the other $Y_i$. In this case, however, we have that $\bigcup\limits_{i} Y_i = Y_k$ which is reflexive by assumption and so $X=Y_k$ means that $X$ is also reflexive.
To make sense out of this question you have to replace the condition $X=\cup_i Y_i$ by the condition that $X$ is the closed subspace generated by reflexive subspaces $Y_i$. But then $X$ need not be reflexive as seen  by considering the collection of all one dimensional subspaces. Of course, there is the trivial case where there is only one $Y_i$ and $X$ is of course, reflexive in that case.