= Solution
Fix a <vertex of a simplicial complex> $v$, and let $R$ be the <set> of vertices reachable from $v$ by an <edge path>. If a <simplex> contains a vertex of $R$, every vertex of that <simplex> lies in $R$: append its <edge> from the reachable vertex to the existing <edge path>. Thus every <simplex> has either all its vertices in $R$ or all outside $R$.
Partition the <geometric realization> as $|K|=A\sqcup B$, according to whether the <support> in the <barycentric coordinate system> lies in $R$ or outside $R$. On each geometric <simplex>, the intersection with $A$ is either the whole <simplex> or the <empty set>, and the same holds for $B$. The defining realization <topology> therefore makes both $A$ and $B$ <closed>, hence also <open> because they are complements.
The vertex $v$ belongs to $A$. If any vertex were unreachable, it would belong to the nonempty set $B$, giving a <separation of a topological space> of $|K|$. This contradicts <connectedness>. Therefore every vertex is reachable from $v$, and any two vertices can be joined by an <edge path>.
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