Showing $[0,1] \times [0,1]$ is a manifold with boundary I'm familiarizing myself with manifolds. I tried to show $[0,1]\times[0,1]$ is a manifold with a boundary. Can you please tell me if my proof is correct: 
The definition for manifold with boundary:
A manifold with boundary $M$ is a second countable Hausdorff space so that for a $p \in M$ there is an open set $U \subseteq M$ so that there is a homeomorphism $\varphi$ to either (a) an open set $V$ in $H^n \setminus \partial H^n$  or to (b) an open set $V$ in $H^n$ and $\varphi (p) \in \partial H^n$ where $H^n$ is the closed upper half plane. It means $H^n = \{x \in \mathbb R^n : x_k \ge 0 ; 1 \le k \le n \}$.
One has to show $M = [0,1] \times [0,1]$ is second countable, Hausdorff and locally homeomorphic to $H^n$.
$M$ is second countable because it has subspace topology of $H^2$.
It is Hausdorff because it has subspace topology of $H^2$.
Locally Euclidean: For point $p \in M$ take set $U$ open in $H^2$ with $p \in U$. Inclusion map $i: U \to U \subseteq H^2$ is local homeomorphism with the property maps to open set and if $p \in \partial M$ then $i(p) \in \partial H^2$.
I am very grateful for your help.
 A: 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.
