2 of 2 improved formatting

Here's a reduction to the finite dimensional case. Let $F$ be a finite set of subspaces of $X$. For each finite dimensional subspace $Y$ of $X$, let $u(Y)$ be the set of elements $Z$ of $F$ such that $Y$ is contained in $Z$. By assumption, $u(Y)$ is non-empty for every $Y$. Since any two finite dimensional subspaces are contained in a third, the intersection of the sets $u(Y)$, as $Y$ runs among all finite dimensional subspaces of $X$, is non-empty. Hence there is at least one set in $F$ that contains every finite dimensional subspace of $X$, hence contains $X$.

For the finite dimensional case, let $F$ be a finite set of subspaces of $X$. By induction, every codimension 1 subspace of $X$ is contained in some $Y$ from $F$. But there are infinitely many codimension $1$ subspaces, so some $Y$ in $F$ contains more than one such subspace. Any two distinct codimension 1 subspaces $\operatorname{span} X$ (if $\dim X > 1$) so $Y = X$.