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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 $|K|=A\bigsqcup B$, 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.