I'm having trouble showing that the Klein bottle defined as a quotient space of $I^2$ with relation $(x,-1)R(x,1)$ and $(-1,y)R(1,-y)$ is Hausdorff and that it can be expressed as $X\cup Y$ where $X,Y$ are homeomorphic to the Möbius strip and $X\cap Y$ is homeomorphic to $S^1$.
[Math] Klein bottle homeomorphic to union of Möbius strip
general-topologyklein-bottlemobius-band
Related Solutions
Imagine the Möbius strip as the unit square, where $(0,x)$ is identified with $(1,1-x)$. The line $\{(x,\frac{1}{2}):x\in[0,1]\}$ is a circle as a subspace of the Möbius strip. Then the map $H:M\times[0,1]\to M$ given by $H((x,y),t)=(x,(1-t)y+\frac{t}{2})$ gives a strong deformation retract of $M$ to the circle.
Below is a picture of the Klein bottle cut into two Möbius strips (along the orange line), so you can see what's going on. To answer your last question: you can apply the above deformation retract to map one of those Möbius bands to the circle. In particular the orange line, which is your $A\cap B$, gets mapped to the circle by this deformation retract.
Each compact surface (except the sphere) has a standard fundamental polygon representation, and each surface belongs to one of the two categories (non-orientable) $mP^2$ or (orientable) $nT^2$. For example, a torus ($T^2$) has a standard representation $aba^{-1}b^{-1}$, see the following figure:
To construct the connected sum between two surfaces, we cut off a corner from each fundamental polygon, then glue the two polygons along the cut off edge. This is a process of concatenating (or inserting) the standard representation of one surface to another. For example, if you glue two torus handles you get a new representation like:
$$aba^{-1}b^{-1}cdc^{-1}d^{-1}$$
This is the standard representation of a $2T^2$, that is, a genus $2$ torus (a torus with two holes).
Back to your question, a Mobius band is a (real) projective space with a disc cut. It has a triangle representation:
Glue the third edge of this triangle with the torus handle, we get a new representation:
$$abcca^{-1}b^{-1}$$
See the following figure:
This representation has an edge pair $cc$ indicating that it belongs to the non-orientable category, and it has a single vertex class thus it has $6$ edges in its standard representation, implying that it is a $3P^2$. That is we've shown that the connected sum between $T^2$ and $P^2$ is
$$T^2\#P^2=3P^2$$
Best Answer
This crude drawing may be helpful.