Showing $[0,1] \times [0,1]$ is a manifold with boundary

The portion where you show that $M$ is locally Euclidean is not correct and here's why:

You have to show that for every point $p \in M$ there exists a neighborhood $U \subseteq M$ of $p$ and a homeomorphism from $U$ to an open subset of either $\mathbb R^2$ or $\mathbb H^2$.

Now here's what you wrote:

• For point $p \in M$ take set $U$ open in $\mathbb H^2$ with $p \in U$

So your neighborhood $U$ is in $\mathbb H^2$ and not in $M$! While it is indeed true that $M \subseteq \mathbb H^2$ we cannot simply say that a neighborhood in $\mathbb H^2$ is a neighborhood in $M$. If $p$ is the point $p = (1, 1)$ then any neighborhood of $p$ in $\mathbb H^2$ will contain points that are not in $M$.

Edit:

To fix the proof remember that if $U \subseteq M$ contains part of the boundry of $M$ then the chart for $U$ will have to map that boundry to the boundry of $\mathbb H^2$. So for example if $p = (1, 1)$ then let $U = (.4, 1] \times (.4, 1]$ and for a chart try $\phi\colon U \to \mathbb H^2$ defined by $\phi(x, y) = (1 - x, 1 - y)$. I leave it to you to check that the image of $\phi$ is open in $\mathbb H^2$ and that $\phi$ is a homeomorphism onto this image. You'll also need to come up with charts covering the rest of $M$. I suggest you take

• $U_2 = [0, .6) \times (.4, 1]$
• $U_3 = (.4, 1] \times [0, .6)$
• $U_4 = [0, .6) \times [0, .6)$

as the neighborhoods for those charts.

Edit #2:

Whoops! My edit above has a mistake as pointed out by goobie. Also pointed out by goobie is the fix: Instead of the chart $\phi$ that I suggested take $\phi\colon U \to \mathbb H^2$ defined by $\phi(z) = z^2$ (here I'm using complex numbers to denote points in the plane). Then you'll just need to do some translation and rotation to handle the other corners.