Here's a reduction to the finite dimensional case. Let F$F$ be a finite set of subspaces of X$X$. For each finite dimensional subspace Y$Y$ of X$X$, let u(Y)$u(Y)$ be the set of elements Z$Z$ of F$F$ such that Y$Y$ is contained in Z$Z$. By assumption, u(Y)$u(Y)$ is non-empty for every Y$Y$. Since any two finite dimensional subspaces are contained in a third, the intersection of the sets u(Y)$u(Y)$, as Y$Y$ runs among all finite dimensional subspaces of X$X$, is non-empty. Hence there is at least one set in F$F$ that contains every finite dimensional subspace of X$X$, hence contains X$X$.
For the finite dimensional case, let F$F$ be a finite set of subspaces of X$X$. By induction, every codimension 1 subspace of X$X$ is contained in some Y$Y$ from F$F$. But there are infinitely many codimension 1$1$ subspaces, so some Y$Y$ in F$F$ contains more than one such subspace. Any two distinct codimension 1 subspaces span X$\operatorname{span} X$ (if dim X > 1$\dim X > 1$) so Y = X$Y = X$.
