Lemma 2

Let X be a complete metric space and let U be an open subset of X. There exists a distance on U compatible with its topology for which U is complete.

We shall take up again the arguments of the proof of prop. 2 of TG, IX, p. 57. We may suppose U distinct from X. Let V be the subset of the product $\mathbf{R}\times X$ formed by the points $(t, x)$ such that $t d(x,X-U) = 1$ ; the subspace V of $\mathbf{R}\times X$ is closed and the mapping $(t, x)\mapsto x$ from V into U is a homeomorphism (TG, IX, p. 13, prop. 3). There exists on $\mathbf{R}\times X$ a distance $d'$ compatible with its topology for which $\mathbf{R}\times X$ is complete (TG, IX, p. 15, cor. 2 and TG, II, p. 17, prop. 10). The space V is complete for the distance induced by $d'$ (TG, II, p. 16, prop. 8), whence the lemma.