It is known that every local homeomorphism $p : X \to Y$ is both continuous and open. Let $f : A \to I, g : B \to I$ be local homeomorphisms and $k : A \to B$ a continuous map be such that $g \circ k = f$. Then $k$ is also an open map.
Let $U \subset A$ be open. Then let $k(x) = y \in k(U) \subset B$ for $x \in U$. Then if $k(U) \supset g^{-1}(V)$ for some open $V \subset I$ with $f(x) \in V$ we're done with the proof.
Let $V = f(U)$. Then $f$ being an open map makes $V$ an open subset of $I$. By commutativity, $g \circ k = f$ we also have that $g\circ k(U) = V$. But I keep seeming to need surjectivity of $g$ to complete the proof.
We can show that the map $k$ is open as follows. Let $O\subset A$ be any open set and $x\in O$ be any point. Pick an open neighborhood $W$ of $f(x)$ such that there exist open neighborhoods $U$ of $x$ and $V$ of $k(x)$ such that both restrictions of $f$ to $U$ and $W$ and of $g$ to $V$ and $W$ are homeomorphisms. Let $h:W\to V$ be a homeomorphism inverse to the latter. Pick an open neighborhood $U’\subset O\cap U$ of $x$ such that $k(U')\subset V$. Then $k(U’)=hf(U’)$ is an open subset of $V$. It follows $k$ is a local homeomorphism too.