If $D$ is an at most countably dimensional division algebra over $\mathbb{C}$ then $D=\mathbb{C}.$
Proof Let $x\in D\setminus\mathbb{C},$ then $\{(x-a)^{-1}, a\in\mathbb{C}\}$ is an uncountable linearly independent set. $\square$
This is an algebraic variant of the Gelfand–Mazur theorem and it implies countably-dimensional Schur's Lemma over $\mathbb{C}$ (or any uncountable field).
