 Solution

The Simplicial approximation theorem states that if $K$ is a finite simplicial complex, $L$ is a simplicial complex, and $f:|K|\to |L|$ is continuous, then for some $r\ge 0$ there is a simplicial approximation $g:K^{(r)}\to L$ to $f$.

The target open stars form an open cover of $|L|$, so their inverse images under $f$ form an open cover of the compact metric space $|K|$. The Lebesgue number lemma supplies $\delta >0$ such that every subset of diameter below $\delta$ lies in one inverse image. Choose $r$ so large that every vertex star in $K^{(r)}$ has diameter below $\delta$; this is possible because its diameter is at most twice $\mu (K^{(r)})$ and the mesh tends to zero.

For each vertex $v$ choose a vertex $w(v)$ of $L$ such thatIf $v_0,\dots ,v_q$ span a simplex, a point in its relative interior belongs to all their open stars. Its image belongs to all the open stars of $w(v_0),\dots ,w(v_q)$, so these target vertices span a simplex. Thus $v\mapsto w(v)$ extends to a simplicial map $g$, and its defining inclusion says exactly that $g$ approximates $f$. This proves the theorem.