This answer benefited from a discussion with I. Izmestiev.
Consider a $3$-dimensional simple polytope $P$. The indegrees of the vertices can only be $0$, $1$, $2$ or $3$. The lowest vertex has indegree $0$, the highest vertex has indegree $3$, and the other vertices have indegree $1$ or $2$. For an index-increasing realization of $P$, there is a plane that separates the indegree-$1$ vertices and the indegree-$2$ vertices.
Moreover, every face of $P$ contains a vertex of indegree $1$ and a vertex of indegree $2$. Therefore, the existence of the required realization means that there must be a plane that cuts all the faces.
From this point of view, it becomes much clearer that index-increasing realization is very unlikely to exist even in dimension $3$. Counter-examples can be easily constructed, for example, by taking truncations.