Lemma 1

Let E be a separated topological vector space of dimension $\geqslant 2$ over $\mathbf{R}$, and let X be a denumerable subset of E. The complementary set E - X is connected.

Suppose first that E is of dimension 2. One may suppose that $E =\mathbf{R}^2$ endowed with the euclidean norm (TVS, I, p. 14, Theorem 2). Since X is denumerable, there exists a real number $r\in \mathbf{R}_+^*$ such that the circle C with center 0 and radius $r$ does not meet X. Let $x\in E$ - X; if $x \notin C$, there exists a point $y\in C$ such that the straight line $L_x$ joining $x$ and $y$ does not meet X, since X is denumerable. The set E - X is the union of C, which is connected, and of the connected sets $L_x\cup C$ for $x\in E$- $(X\cup C)$; these sets all contain C, and therefore E- X is connected (TG, I, p. 81, Proposition 2).

Consider the general case. Replacing X by $X-x$ for an element $x\in E$ - X, one is reduced to the case where $0\notin X$. Since E - X is the union of the sets F - $(X\cap F)$ when F runs through the set of 2-dimensional subspaces of E, and since these are connected by the preceding case and contain 0, the set E - X is connected (loc. cit.).