= Solution
A <boundary> reparameterization gives an explicit <homeomorphism>. Let $Q=[0,1]^2$, with centre $c=(1/2,1/2)$, and parametrize its <boundary> once by
$$
b(s)=\begin{cases}(s,1),&0\leq s\leq1,\\(1,2-s),&1\leq s\leq2,\\(3-s,0),&2\leq s\leq3,\\(0,s-3),&3\leq s\leq4.\end{cases}
$$
Here $b(0)=b(4)$. The omitted top edge of $[0,1]\times[0,1)$ is $b([0,1])$. The omitted top and right edges of $[0,1)^2$ are $b([0,2])$.
Use the increasing <boundary> parameter map
$$
\phi(s)=\begin{cases}2s,&0\leq s\leq1,\\2+\frac23(s-1),&1\leq s\leq4.\end{cases}
$$
It takes $[0,4]$ bijectively onto itself, respects the identified endpoints, and maps the first omitted arc exactly onto the second. Extend it radially:
$$
\boxed{H(c)=c,\qquad H\bigl(c+r(b(s)-c)\bigr)=c+r\bigl(b(\phi(s))-c\bigr),\quad0<r\leq1.}
$$
Every noncentral point has a unique <boundary> ray and a unique radial parameter $r=2\|q-c\|_\infty$; the only duplicate <boundary> parameters are zero and four, whose images agree. The inverse uses $\phi^{-1}$ by the same construction. Both maps are <continuous> off the centre, and at the centre their radial distances tend to zero. Thus $H$ is a <homeomorphism> of the whole <closed> square.
It carries the first omitted edge precisely onto the second omitted pair of edges, so restriction gives
$$
\boxed{H:[0,1]\times[0,1)\ \cong\ [0,1)\times[0,1).}
$$
This is the <radial extension of a square boundary homeomorphism>; it avoids any need to treat corners as topologically distinguished points.
\Image[/past-exam-of-the-topology-course-of-the-university-of-oxford/2014-half-open-squares.png]
{title=A boundary reparameterization joins two omitted edges into the image of one}
{description=Red boundary arcs are omitted from the respective spaces. Corresponding rays from the centre show the radial extension; the two blue ray families are related by the explicit piecewise linear boundary parameter map.}
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