Q:

question regarding polyhedra in algebraic topology

This picture is taken from algebraic topology  by Rotman.  I have two question regarding this section. 
(1) How to prove Real projective plane is a polyhedra.  Although they have said it will lead to polyhedra but didn't give any solution for it. 
(2) Most Importantly how can I prove excercise 7.1, I have been seeing many people posting the solution although assume $RP^2$ is obtained after identifying  antipodal points of $S^2$, although I dont know how to prove that either.  So it will be great if you can provide me a proof or a hint of the question. 

A:

I'll post them here because the OP requested an answer. 

Is it a polyhedron? 

Yes, by Paul Frost's comment below the post (I don't have Rotman's book)
to see this, you could split the rectangle into 9 retangular regions and subdivide each region into two triangles by joining the two long diagnal lines and two "medium-sized" diagnal lines, which would give a valid triangulation. 
(just corrected a terrible error: Only one diagnal line doesn't work because the image of each simplex won't be spanned by three vertices, after identification)

Prove $X=I^2/{\sim}=[-1,1]^2/{\sim}=\Bbb{RP}^2$ is homeomorphic to $D^2/{x\sim-x}$ for $x\in\partial D^2\approx S^1$.

For the current boundary of the representation of $\Bbb{RP}^2$, we can see that it is homeomorphic to $S^1/{\sim}$ by $f:\partial X\to \partial D^2/{\sim}, (x',y')\mapsto\frac{(x',y')}{\sqrt{x'^2+y'^2}}$ because it is a continuous bijection from a compact space to a hausdorff space, where $\partial X$ refers to the boundary of the square representation of $\Bbb{RP}^2$. Therefore $f^{-1}$, without explicitly writing it out, is also a homeomorphism. 
Define $h:D^2/{\sim}\to X$, an extension of $f^{-1}$ to the interior of the disk:
$$
h(x,y)=
\begin{cases}
(0,0) & (x,y)=(0,0)\\
\sqrt{x'^2+y'^2}(x,y) & \text{otherwise}
\end{cases}
$$
It is clearly bijective, and it's continuous on the second branch.  To verify the continuity at $(0,0)$, consider $\lim_{(x,y)\to(0,0)}\sqrt{x'^2+y'^2}(x,y)=(0,0)$. Thus, the first branch is a continuous extension of the second which implies $h$'s continuity.  Because $h$ is a continuous bijection from a compact space to a hausdorff space and is equivalence relation preserving, it's a homeomorphism. 
A simple remark: I didn't combine the two branches is because $\sqrt{x'^2+y'^2}=||f^{-1}(\frac{(x',y')}{\sqrt{x'^2+y'^2}})||$, which is undefined at $(0,0)$, so we have to prove the continuity of the extension just like what we did in the preceding paragraph. 
Edit:
To visualize this homeomorphism, I found that this picture from Munkres's Topology might work as well:

Now, you could easily see that any point on the boundary of the 2-disk is identified with its antipodal point by the labelling scheme. 
To check that the equivalence classes are preserved, we can do the following:
In $X$, $(x,-1)\sim(-x,1)$, $f(x,-1)$ ends up with an arc on the unit circle by $f:(x',y')\mapsto\frac{(x',y')}{||(x',y')||}$. Now, by direct calculation, $f(x,-1)$ lies between $(-\sqrt{2}/2,\sqrt{2}/2)$ and $(\sqrt{2}/2,-\sqrt{2}/2)$, which can be represented as $$f(x,-1)=(\cos([\dfrac{5}{4}\pi,\dfrac{7}{5}\pi]),\sin([\dfrac{5}{4}\pi,\dfrac{7}{4}\pi]))$$
$f(-x,1)$ lies between $f(-1,1)=(-\sqrt{2}/2,\sqrt{2}/2)$ and $f(1,1)=(\sqrt{2}/2,\sqrt{2}/2)$, which can be represented as $$(\cos([\dfrac{1}{4}\pi,\dfrac{3}{5}\pi]),\sin([\dfrac{1}{4}\pi,\dfrac{3}{4}\pi]))=(\cos([\dfrac{5}{4}\pi-\pi,\dfrac{7}{5}\pi-\pi]),\sin([\dfrac{5}{4}\pi-\pi,\dfrac{7}{4}\pi-\pi]))$$
Here, trigonometry tells us that they're pairs of antipodal points.  A similar argument shows that the same thing happens on $(-1,y)\sim(1,-y)$. $h$ also satisfies this because $h|_{\partial D^2}=f^{-1}$, which sends everything back by a continuous bijection. 

Any question please feel free to ask.