How to see the real projective plane is a Möbius band glued to a disk?

I am seeking an easily comprehended, convincing explanation that ${RP}^2$ is topologically the same as gluing the circle boundary of a disk to the edge of a Möbius band.

Let $D$ be the closed unit disk in $\mathbb{R}^2$, and $D/\sim$ the disk with antipodal points on the boundary identified, which is homeomorphic to $\mathbb{RP}^2$.

Now decompose $D$ into an annulus $A$ and a smaller disk, so that attaching a disk to $A$ along the inner circle gives you $D$.

So, attaching a disk to $A/ \sim$ along the inner circle will give you $(D/\sim) \cong \mathbb{RP}^2$. If we can show that $A/\sim$ is homeomorphic to a Möbius band, we're done.

Here's how we do that.

(The image is from the Oxford Part A Topology lecture notes)

It's really the same thing as Daniel's answer, but I find this easier to visualize:

Consider $\mathbb{RP}^2$ as a quotient of $S^2$. The image of the disk $\{(x,y,z)\in S^2\mid z\geq1/2\}$ is the disk, and the image of the strip $\{(x,y,z)\in S^2\mid y\geq0,\,|z|\leq1/2\}$ is the Möbius strip.