This question is motivated by Jyrki Lahtonen's comment in this question. It was an attempt at a short proof that for arbitrary field $k$, we have that $k^\mathbf N$ (as the full set-theoretic product with obvious linear structure) is not of countable dimension over $k$.
Don't know about one-liners. You could also argue that the subspaces $V_r$ defined for all real numbers $r$ as $V_r:=\{ s:\mathbf Q \to k \lvert s(q)=0\forall q>r\}$ form an uncountable chain of subspaces (without repetitions): $V_{r_1}\subsetneq V_{r_2}$ whenever $r_1
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At first glance, it seems like a really nice and elegant argument, but when you look a little closer, there's no obvious (to me) reason for it not to work in the case of a countably infinite dimensional space. Sure, individual $V_r$ may be smaller, but they still seem to form an uncountable strictly increasing chain of subspaces, contradicting countable dimension.
This obviously can't be right, so where does the problem lie?
Countability is probably immaterial here, you could make a very similar argument using some linear orders larger than rationals, dense in yet larger linear orders to obtain pretty much the same "result" for arbitrarily large cardinals.