This following gives an example, probably:
Let $H=\{(x,y)\in\mathbb R^2:y>0\}$ be the upper half plane in $\mathbb R^2$ and let $\bar H$ be its closure. Let me give you an homeo $f:H\to H$ which does not extend to an homeo $\bar f:\bar H\to\bar H$, and which fixes the point of infinity on the $x$-axis. Then you can conjugate with an homeo $D\to H$ from the open unit disk which extends to the boundary (apart from one point...)
Let $\phi:\mathbb R\times(0,\infty)\to\mathbb R$ be given by $\phi(x,t)=\frac1{\sqrt{4\pi t}}\exp\left(-\frac{x^2}{4t}\right)$, and let $h:\mathbb R\to\mathbb R$ be a positive smooth function with support on $[-2,2]$ and such that it is constantly $1$ on $[-1,1]$. Define now $g:H\to H$ so that when $x\in\mathbb R$ and $y>0$ we have $$
g(x,y)=\int_{-\infty}^\infty\phi(x-\xi,y)(1-h(\xi))\,\mathrm d\xi.
$$ This is a solution of the one-dimensional head equation. Properties of that equation (which one can easily show in this case!) imply that $f$ extends continuously to a map $\bar g:\bar H\to\mathbb H$.
Now consider the map $f:(x,t)\in H\mapsto (xg(x,t),t)\in H$. This is an homeomorphism, and it extends to the map $\bar f:(x,t)\in\bar H\mapsto (x\bar g(x,t),t)\in\bar H$, which is not an homeo.