Here is a proof which uses only singular homology.$\newcommand{\RR}{\mathbb{R}}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\To}{\longrightarrow}
\newcommand{\set}[1]{\lbrace #1 \rbrace}
\newcommand{\Xminusx}{X\setminus\set{x}}
\newcommand{\Yminusy}{Y\setminus\set{y}}$

Assume $f:X\times Y\to\RR^n$ is a homeomorphism, and that $X$ is compact. I will prove that $X$ is a singleton, by repeatedly applying the Künneth theorem.


### General remarks ###

The spaces $X$ and $Y$ cannot be empty, and we will fix $x\in X$ and $y\in Y$. Observe that $X$ and $Y$ are Hausdorff, given that $\RR^n$ is Hausdorff. In particular, $\Xminusx$ is open in $X$, and $\Yminusy$ is open in $Y$. Let also $p=f(x,y)\in\RR^n$. Finally, as observed in the comments, $X$ and $Y$ are contractible since $\RR^n$ is contractible. In particular, $H_\ast(X)$ is zero in positive degrees, and $\ZZ$ in degree zero.


### Claim 1: $H_n(Y,\Yminusy) \simeq H_n\bigl(\RR^n,\RR^n\setminus f(X\times\set{y})\bigr)$ ###

First of all, consider the pair $X\times(Y,\Yminusy)=\bigl(X\times Y,X\times(\Yminusy)\bigr)$. By the Künneth theorem and the contractibility of $X$, we conclude that
$$ H_n\bigl(X\times Y,X\times (\Yminusy)\bigr) = H_0(X)\otimes H_n(Y,\Yminusy) = H_n(Y,\Yminusy) $$
The above pair $\bigl(X\times Y,X\times (\Yminusy)\bigr)$ is homeomorphic via $f$ to $\bigl(\RR^n,\RR^n\setminus f(X\times\set{y})\bigr)$. The preceding expression thus implies
$$ H_n\bigl(\RR^n,\RR^n\setminus f(X\times\set{y})\bigr) \simeq H_n(Y,\Yminusy) $$


### Claim 2: $H_n(Y,\Yminusy)$ has a $\ZZ$ summand ###

Since $X$ is compact, the image $f(X\times\set{y})$ is compact in $\RR^n$, and thus bounded. Let $R\in\RR^+$ be such that $f(X\times\set{y})$ is contained in the closed ball of radius $R$ centered at $p$, $B_R(p)$. Then we have inclusions
$$ \RR^n\setminus B_R(p) \subset \RR^n\setminus f(X\times\set{y}) \subset \RR^n\setminus\set{p} $$
which induce homomorphisms on homology:
$$ \ZZ \simeq H_n(\RR^n,\RR^n\setminus\set{p}) \To H_n\bigl(\RR^n,\RR^n\setminus f(X\times\set{y})\bigr) \To H_n\bigl(\RR^n,\RR^n\setminus B_R(p)\bigr) \simeq \ZZ $$
The composition of the two maps is an isomorphism, therefore they exhibit a splitting of the middle group:
$$ H_n(Y,\Yminusy) \simeq H_n\bigl(\RR^n,\RR^n\setminus f(X\times\set{y})\bigr) \simeq \ZZ \oplus A $$
for some abelian group $A$.


### Claim 3: $H_\ast(X,\Xminusx)$ is concentrated in degree zero ###

Let $i$ be a positive integer. Observe that $f$ gives a homeomorphism between the pairs $\bigl(X\times Y,(X\times Y)\setminus\set{(x,y)}\bigr)$ and $(\RR^n,\RR^n\setminus\set{p})$. Consequently,
$$ H_{n+i}\bigl(X\times Y,(X\times Y)\setminus\set{(x,y)}\bigr) \simeq H_{n+i}(\RR^n,\RR^n\setminus\set{p}) = 0 $$

Recall that $\Xminusx$ and $\Yminusy$ are open in $X$ and $Y$, respectively. So we can apply the Künneth theorem to the pair
$$ (X,\Xminusx)\times(Y,\Yminusy) = \bigl(X\times Y,(X\times Y)\setminus\set{(x,y)}\bigr) $$
which implies that there is a monomorphism
$$ H_i(X,\Xminusx)\otimes H_n(Y,\Yminusy) \To H_{n+i}\bigl(X\times Y,(X\times Y)\setminus\set{(x,y)}\bigr) = 0 $$
It follows that $H_i(X,\Xminusx)\otimes H_n(Y,\Yminusy) = 0$. Since $H_n(Y,\Yminusy)$ contains a summand isomorphic to $\ZZ$, we conclude that $H_i(X,\Xminusx)=0$.


### Claim 4: $H_0(X,\Xminusx)$ is not zero ###

We now know that $H_\ast(X,\Xminusx)$ is zero in positive degrees, and it is necessarily a free abelian group in degree zero. Applying once more the Kunneth theorem to $(X,\Xminusx)\times(Y,\Yminusy)$, we obtain an isomorphism
$$\begin{array}{rl}
H_0(X,\Xminusx)\otimes H_n(Y,\Yminusy) \!\!\!\! & = H_n\bigl(X\times Y,(X\times Y)\setminus\set{(x,y)}\bigr) \\
& \simeq H_n(\RR^n,\RR^n\setminus\set{p}) \\
& \simeq \ZZ 
\end{array}$$
Consequently, $H_0(X,\Xminusx) \neq 0$.

### Conclusion ###

Since $H_0(X)=\ZZ$, the only way that we can have $H_0(X,\Xminusx) \neq 0$ is if $\Xminusx = \emptyset$. Thus $X=\set{x}$ is a singleton.