So to show this is a 2 manifold with boundary you have to show that around each point there is a neighborhood that is either homeomorphic to $D^2$ or $D^2_+= \{(x,y)\in \mathbb{R} | \,\,\,\, y\geq 0, \,\,\,\, |(x,y)|<1 \}$.
Let $X$ be the described set $X / \sim$ the quotient and $\pi$ the quotient homomorphisim.
For $x \in \pi( \text{int} \, ( X )) = \text{int} \, (X)$ we are done, this set is homeomorphic to the disk. On $\text{int} \,(X)$, $\pi$ is a homeomorphism.
For $x \in \pi( (-10, 10) \times \{1\})$ consider $\pi((-10, 10) \times [1,-1))$. Similarly for the other side.
For $x \in \pi( \{10\} \times (-1,1) )$ it is more difficult. Here we have to somehow work with the twist. Let $f: [-10,-9) \cup (9,10] \times (-1,1) / \sim \,\, \to (-1,1)^2 $ be:
$$
f(x,y) = \left\{
\begin{array}{lr}
(x-10,y) & : x \in (9,10] \\
(x+10,-y) & : x \in [-10,-9)
\end{array}
\right.
$$
I claim that this is continuous and bijective. Pulling $f$ back to $X$, i.e. considering $f \circ \pi : X \to (-1,1)^2$, it is continuous (this is the universal property of quotients). And it is bijective as $f \circ \pi$ is 1 to 1 except for the points that are identified where is is 2 to 1. But those points are identified so $f$ is 1 to 1 and onto. $f$ is also an open map, any open set in $X / \sim$ is the union of the images of an open sets from $X$ and $f \circ \pi$ is clearly an open map.
The natural way to show this would be to explicitly write down the embedding.
For example, the Möbius strip can map to a thin belt around the equator of the sphere. Then the relation between this belt (in spherical coordinates!) and your defining parameterization of the Möbius strip is a simple affine transformation of each coordinate separately. Remember that $[0,1]$ should map to only one hemisphere of longitude due to the identification of antipodes.
Then there remains some slightly tedious fiddlework proving that your map respects the two relations you quotient out, and you're set.
But actually there's no reason to make it a thin belt. If instead you make the belt extend all the way to the poles, you'll have proved the stronger result that the real projective plane minus one point is homeomorphic to the Möbius strip without boundary.
Best Answer
I'll try to answer the intuition part. You know how a Möbius strip works with a piece of paper, right?
Visualize $\mathbb{R}\times [-1,1]$. This is the whole real line in the $x$-direction, and the interval $[-1,1]$.
Now, what parts of this does $\sim$ say are the same? When $y=0$, every $x$ is the same as every $x+n$ for any $n\in \mathbb{N}$. Now, move up a little bit in the $y$-direction, to some $a\in (0,1)$. Now any $(x,a)$ is identified with $(x+1,-a)$. So, if you just look at $A_1:=[-1,0]\times [-1,1]$ and $A_2:=[0,1]\times [-1,1]$, you see that the top of $A_1$ is identified with the bottom of $A_2$.
In fact, every $(x,y)$ is identified with $(x+1,-y)$, $(x+2,y)$, $(x+3,-y)$, etc. The top of each interval is identified with the bottom of its successor's. We can move down a path like this through points which will later be identified with one another - in the quotient, we would just be going through the same one interval long path over and over.
Before we take the quotient, the path going through the tops of the even intervals and the path going through the top of the odd intervals are different.
Here we can see how the paper Möbius strip corresponds to the mathematical one; before you twist and glue, it starts off with two sides. (EDIT: Don't be fooled by the ends of my sticky notes; this strip is supposed to be one interval long.)
Once we quotient out by $\sim$, we pull these identified points together according to the orientation given by this equivalence relation. The red and green paths show how the orientation 'flips' at the end of every interval. We are left with the Möbius strip.