I am having major trouble showing that the version of the projective plane here (with a Mobius strip) is homeomorphic to the projective plane that is defined as the quotient of the sphere $S^2$..
Is this a matter of showing one is orientable and the other isn't (the one with the Mobius Strip) and going from there? Not sure what to do. Thanks for any help/attempts at a proof.
First, notice that in the diagram you provided, if you are given two points $p \ne q$ in the square, those two points $p,q$ are identified under the gluing if and only if $p,q$ are endpoints of a "diameter" of the square, meaning that the line segment $\overline{pq}$ passes through the center of the square.
Now round out the square to get a round disc $D^2 = \{(x,y) \mid x^2 + y^2 \le 1\}$. It's boundary circle is $\partial D^2 = \{(x,y) \mid x^2 + y^2 = 1$. Again the identifications are that two points $p \ne q \in D^2$ are identified if and only if $\overline{pq}$ is a diameter of $D^2$.
Now consider the upper hemisphere of $S^2$, namely $$S^2_+ = \{(x,y,\sqrt{1-x^2-y^2}) \mid x^2 + y^2 \le 1\} $$ The boundary of $S^2_+$ is the equator of $S^2$. Again two points $p \ne q \in S^2_+$ are identified if and only if $p,q$ endpoints of a diameter of the equator.
Finally, take the union of the upper and lower hemispheres to get the whole sphere $S^2$, where two points $p \ne q \in S^2$ are identified if and only if they are endpoints of a diameter of $S^2$. You get a bijection between the quotient of $S^2_+$ and the quotient of $S^2$, because if $p \ne q \in S^2_+$ then they are identified in the quotient of $S^2_+$ if and only if they are identified in the quotient of $S^2$.
That should give you enough intuition. There are some further formalities that require some topology, to prove that this sequence of maps between quotients is a sequence of homeomorphisms.