I couldn't find the definition of a simple poset, but I think the following should count as a counterexample. Let $G$ be the edge graph of the octahedron, so $G$ has $6$ vertices which I'll label by $V:=\{ -3,-2,-1,1,2,3 \}$ and has as edges all pairs $(i,j)$ from $V$ except those of the form $(i, -i)$. So $G$ is $4$-regular. I claim that

• $G$ is not the edge graph of a $4$-polytope but

• There is a regular CW decomposition of the $3$-sphere with edge graph $G$.

Verification that $G$ is not the edge graph of a $4$-polytope. Consider what the $3$-faces of sch a polytope can be. Any $3$-polytope must have at least $4$-vertices, and if it has exactly $4$ then it is a tetrahedron, so its edge graph is a $K_4$. But $G$ contains no $K_4$, so none of its $4$-element subsets can be the vertices of a $3$-face. If all $6$ vertices were the vertices of a $3$-face, then we would have a $3$-dimensional polytope, not a $4$-dimensional one.

So all faces must use $5$ vertices. If $V \setminus \{ i \}$ is a face, then it must be a square pyramid, with $-i$ at the vertex.

Now, suppose that $V \setminus \{ i \}$ and $V \setminus \{ j \}$ are both faces. I claim that we must have $i = -j$. Suppose otherwise. Then the vertices of $V \setminus \{ i \}$ lie in an affine $3$-plane. Moreover, since $j$ is NOT the apex of the pyramid $V \setminus \{ i \}$, this $3$-plane is affinely spanned by $V \setminus \{ i,j \}$. So vertex $i$ is in the $3$-plane spanned by $V \setminus \{ i,j \}$. But the same holds for vertex $j$. So all $6$ vertices would be in an affine $3$-plane, a contradiction.

We have now shown that there are at most two $3$-faces in our polytope, which is absurd.

Construction of a regular CW decomposition of $S^3$ with edge graph $G$.

Take four square pyramids, with vertex sets $(1, \pm 2, \pm 3)$, $(-1, \pm 2, \pm 3)$, $(\pm 1, 2, \pm 3)$ and $(\pm 1, -2, \pm 3)$, and with $1$, $-1$, $2$ and $-2$ at their apices respectively. Gluing the first two together along their square faces gives a $3$-ball with boundary the octahedron; gluing the latter two along their square faces gives another $3$-ball with boundary the octahedron; now glue the two octahedra together to give an $S^3$.

This is the face structure of the totally nonnegative part of the Grassmannian $G(2,4)$.

I couldn't find the definition of a simple poset, but I think the following should count as a counterexample. Let $G$ be the edge graph of the octahedron, so $G$ has $6$ vertices which I'll label by $V:=\{ -3,-2,-1,1,2,3 \}$ and has as edges all pairs $(i,j)$ from $V$ except those of the form $(i, -i)$. So $G$ is $4$-regular. I claim that

• $G$ is not the edge graph of a $4$-polytope but

• There is a regular CW decomposition of the $3$-sphere with edge graph $G$.

Verification that $G$ is not the edge graph of a $4$-polytope. Consider what the $3$-faces of sch a polytope can be. Any $3$-polytope must have at least four vertices, and if it has exactly four then it is a tetrahedron, so its edge graph is a $K_4$. But $G$ contains no $K_4$, so none of its $4$-element subsets can be the vertices of a $3$-face. If all six vertices were the vertices of a $3$-face, then we would have a $3$-dimensional polytope, not a $4$-dimensional one.

So all faces must use five vertices. If $V \setminus \{ i \}$ is a face, then it must be a square pyramid, with $-i$ at the apex.

Now, suppose that $V \setminus \{ i \}$ and $V \setminus \{ j \}$ are both faces. I claim that we must have $i = -j$. Suppose otherwise. Then the vertices of $V \setminus \{ i \}$ lie in an affine $3$-plane. Moreover, since $j$ is NOT the apex of the pyramid $V \setminus \{ i \}$, this $3$-plane is affinely spanned by $V \setminus \{ i,j \}$. So vertex $i$ is in the $3$-plane spanned by $V \setminus \{ i,j \}$. But the same holds for vertex $j$. So all six vertices would be in an affine $3$-plane, a contradiction.

We have now shown that there are at most two $3$-faces in our polytope, which is absurd.

Construction of a regular CW decomposition of $S^3$ with edge graph $G$.

Take four square pyramids, with vertex sets $(1, \pm 2, \pm 3)$, $(-1, \pm 2, \pm 3)$, $(\pm 1, 2, \pm 3)$ and $(\pm 1, -2, \pm 3)$, and with $1$, $-1$, $2$ and $-2$ at their apices respectively. Gluing the first two together along their square faces gives a $3$-ball with boundary the octahedron; gluing the latter two along their square faces gives another $3$-ball with boundary the octahedron; now glue the two octahedra together to give an $S^3$.

This is the face structure of the totally nonnegative part of the Grassmannian $G(2,4)$.