Let $A$ and $B$ be path connected open subspaces of a topological space $X$ and assume that $A\cap B\neq \emptyset$ is simply-connected. Let $x$ and $y$ be two points in $A\cap B$.
Let $\gamma$ and $\theta$ be paths in $A$ and $B$ respectively, both of which start at $x$ and ends at $y$.
Assume that there is a path homotopy $H:I\times I\to X$ between $\gamma$ and $\theta$, that is, $[\gamma]=[\theta]$ in $X$.
Question. Is is necessarily true that there is a path $\alpha$ in $A\cap B$ which starts at $x$ and ends in $y$, such that $\alpha$ is path homotopic to $\gamma$ in $A$?
The intuition behind the above question is that as $\gamma$ is deformed by $H$, each point of $\gamma$ should pass through $A\cap B$ at some point of time. In fact, by Lebesgue Number Lemma, the is a subdivision of $\gamma$ such that each part of the subdivision is mapped into $A\cap B$ by $H$ for the same time $t$. Now since $A\cap B$ simply connected, we should be able to connect these chunks to get a path that lies entirely in $A\cap B$.
EDIT:
Here is what I was trying to say in my 'Lebesgue Number Lemma' line above.
Claim. There is a subdivision $0=s_0< s_1\cdots <s_n=1$ of $[0, 1]$ such that for each $0\leq i\leq n-1$, there exists $t_i\in I$ satisfying $H([s_i, s_{i+1}]\times \{t_i\})\subseteq A\cap B$.
Proof. For each $s\in I$, there is a time $t_s\in I$ such that $H(s, t_s)\in A\cap B$. Since $A\cap B$ is open, for each $s\in I$ there is a neighborhood $U_s$ of $s$ in $I$ such that $H(U_s\times \{t_s\})\subseteq A\cap B$. Now $\{U_s\}_{s\in I}$ forms an open cover in $I$ and therefore admits a Lebesgue number. Choose a subdivision $0=s_0< s_1\cdots <s_n=1$ of $I$ such that each $s_{i+1}-s_i$ is smaller than this Lebesgue number, so that for each $i$ we have a $t_i$ such that $H([s_i, s_{i+1}]\times \{t_i\})\subseteq A\cap B$.
What I am going to write is essentially a simple rewrite of Hatcher's proof of van Kampen's theorem for your case, and without all the details (I can provide them if you want), which I hope it is clearer.
Consider the homotopy $ H \colon I \times I \to X$ of paths from $\gamma$ to $\theta$. By a Lebesgue number lemma argument, there are partitions $ 0 = s_0 < s_1 < \ldots < s_n = 1$ and $ 0 = t_0 < t_1 < \ldots < t_m = 1$ of $I$ such that each $H([s_i,s_{i+1}] \times [t_j,t_{j+1}])$ is contained in $A$ or in $B$. Since the top is contained in $B$ and the bottom is contained in $A$, we can do this in such a way that the rectangles in the lowest row are mapped to $A$ and the rectangles in the highest row to $B$. I am going to write a diagram for it in this way, assumming $ m = n =4$ for simplicity and also a very special situation.
$$ \begin{array}{|c|c|c|c|} \hline B & B & B & B \\ \hline B & A & A & B \\ \hline A & A & A & B \\ \hline A & A & A & A \\ \hline \end{array} $$
Consider now the path in $I \times I$ that goes from $(0,0)$ to $(0,2)$, then to $(1,2)$, then to $(1,3)$, then to $(3,3)$, then to $(3,1)$, then to $(4,1)$ and finally to $(4,0)$. (Sorry, I don't know how to just color it in the diagram). The image of this path under $H$ is a path $\alpha \colon I \to A \cap B$.
And the higher part of the diagram shows that $\alpha$ and $\theta$ are path homotopic in $B$, while the lower part shows that $\alpha$ and $\gamma$ are path homotopic in $A$.
Of course, this is not necessarily this easy, the diagram could have been
$$ \begin{array}{|c|c|c|c|} \hline B & B & B & B \\ \hline B & A & A & B \\ \hline A & B & A & B \\ \hline A & A & A & A \\ \hline \end{array} $$
We would again take the same path anyway.
Now here is the trick, you can change the homotopy on the little square with is bothering us so that its image falls in $A$. To do that, consider that this square is surrounded by $A$, so the sides are all in $A \cap B$. The images under $H$ of the path that goes from $(1,1)$ from $(2,1)$ and the path that goes from $(1,1)$ to $(1,2)$, then to $(2,2)$ and then to $(2,1)$ are path homotopic, because $A \cap B$ is simply connected. We can use this homotopy to change $H$ on that square so that the image falls in $A \cap B$ without changing on the boundary of the square so that the resulting homotopy is continuous. In particular, the image falls in $A$ and we are in our previous situation.
In general, if we choose the path that goes along the higher part of the highest $A$'s in each column and the sides, all the "blocks" (not necessarily a square, but a few squares togethers) of $B$'s below will be surrounded by $A$ and the same argument can be repeated. In a finite number of steps, you are done.