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.
Note that if you cut a disk out of your lens shape, the remaining annulus is embedded in $\mathbb{R}^3$ without any "twists". If this could be "glued" to itself to form a Möbius band in $\mathbb{R}^3$, then the (external) boundary of your lens-shape region would be mapped to the "soul" of the Möbius band; namely, the curve running along the middle of the Möbius band for one of the standard embeddings.
However, it is easy to check that if one cuts the Möbius band open along its "soul", one gets a strip with two twists as you make full circle around it, rather than the strip with no twists that would result if one removed a disk from the "lens". Thus such a construction is impossible. Note that the use of the term "soul" is consistent with the term used in Riemannian geometry for manifolds of nonnegative curvature.
Best Answer
I've attached a model for the Möbius band. The vertices are $a$ and $b$. The ones that eventually get glued together are given the same letter. The edges are labeled $A$, $B$ and $C$. There are two edges labelled $A$, and they have arrows. The must be glued so that the arrows agree, i.e. you need to give a half twist before you glue. The arrows on $B$ and $C$ are only there because we must orient simplices. Finally, $\alpha$ is the one face. I don't know how to put a clockwise arrow around it, so please add one.
To find the homology groups, we must look at the images and the kernels of the boundary maps. Consider the series of maps $0 \to F \to E \to V \to 0$, where $F$ stands for faces, $E$ for edges and $V$ for vertices. In between each is a boundary map.
We can put all of this together. The group $H_2(M,\mathbb{Z})$ is given by the quotient of the kernel of $F \to E$ by the image of $0 \to F$, i.e. $0/0 \cong 0$. The group $H_1(M,\mathbb{Z})$ is given by the quotient of the kernel of $E \to V$ by the image of $F \to E$, i.e. $\mathbb{Z}^2/\mathbb{Z} \cong \mathbb{Z}$. The group $H_0(M,\mathbb{Z})$ is given by the quotient of the kernel of $V \to 0$ by the image of $E \to V$, i.e. $\mathbb{Z}^2/\mathbb{Z} \cong \mathbb{Z}$. Hence:
\begin{array}{ccc} H_2(M,\mathbb{Z}) &\cong& \{0\} \\ H_1(M,\mathbb{Z}) &\cong& \mathbb{Z} \\ H_0(M,\mathbb{Z}) &\cong& \mathbb{Z} \end{array}