Suppose that $X$ is a real separable vector space, and $W$ a closed linear subspace of $X$. Show that there exists a sequence $(z_j )\in X$ such that $z_{j+1} \notin W_j := \mathrm{span} \ W \cup \{z_1, . . . , z_j\}$ and if we define $W_\infty = \mathrm{span} \ W \cup \{z_j\}_{j=1}^\infty$ then $\overline{W_\infty} = X$.
This is a problem in a (non-credit) example sheet I am working on currently. After spending some time trying it I have realised that it is probably incorrect as it stands, but I am not sure. I think that (even with the necessary assumption that $W$ is a proper subspace), $W=\mathbb{C}$ (over $\mathbb{R}$) is a counterexample? Maybe there is some implicit assumption in the problem that I am missing. I would appreciate if someone could clarify what the correct version of the problem would be. Please do not give me answers to the problem though, as I would like to first try it myself.
As stated, the result is false even for infinite-dimensional $X$. Because if for instance the codimension of $W$ is finite (say the kernel of a bounded linear functional, which has codimension 1), then no such sequence exists.
The easiest way to make the problem solvable would be to take $W$ finite-dimensional (of course you need $X$ infinite-dimensional, as the sequence $\{z_j\}$ is linearly independent). But that's clearly not the intention, since then it would irrelevant to say that $W$ is closed (as all finite-dimensional subspaces are). The other option I see is to say out right that the codimension of $W$ is infinite.