Question of whether two given spaces are homeomorphic. Let $D^2$ be the closed disk on the plane. 
First we pick an arbitrary point $x\in bd(D^2)$ on $D^2$, and define $X = D^2-\{x\}$.
Then define another space $Y$ by removing a homeomorphic image of the closed interval $I$ from the boundary, that is, $Y = D^2-h(I)$.(The original question defines $Y$ by removing the upper closed semi-circle of $D^2$ while I think this can be generalized.)
I tried to construct the homeomorphism between two spaces but failed. The two spaces are both connected, non-compact and convex so I also failed to prove they are not homeomorphic. 

The answer of @PaulFrost is awesome, and should be the accepted answer. I thought I'd just give an explicit description of his approach.
We'll write the unit disk as 
$$ \mathbb{D} = \{re^{\pi it}\mid 0\le r\le1,\ -1\le t\le1\}$$
And define the lower hemisphere as $H=\{e^{\pi it}\mid -1\le t\le0\}$, and the point $q=-1$. 
We have a continuous map $h:\mathbb{D}\rightarrow\mathbb{D}$, given by
$$ h(re^{\pi it}) = re^{t-r+r|t|}$$
It should be checked this is well-defined (plugging in $t=1$ and $t=-1$). It's also straightforward that $h^{-1}(q)=H$. Thus we get a map $h:\mathbb{D}\setminus H\rightarrow \mathbb{D}\setminus\{q\}$. 
An explicit inverse can be given too:
$$ h^{-1}(\rho e^{\pi i\theta}) = \rho e^{(\rho+\theta)^2/(\rho+\theta+\rho|\rho+\theta|)}$$
One has to worry about that denominator being zero. The quadratic formula shows that happens when $\rho+\theta=0$. If $\rho<1$, then $-1<\theta<0$, and it is easy to check the fraction in the exponent goes to zero. So $h^{-1}(\rho e^{-\pi i\rho})=\rho$.
When $\rho=1$, that's a prpblem, because then we have $\theta=\pm1$. Luckily we are ignoring that point ($q$), so we get a well-defined map
$$ h^{-1}:\mathbb{D}\setminus \{q\}\rightarrow\mathbb{D}\setminus H$$
and so $h$ is the desired homeomorphism. 
