We need $k$ to be an infinite field. If $k$ is finite, since point sets are closed in the Zariski topology, we could find two proper closed sets (the union of all but one point of $X$) whose union is all of $X$.
Suppose $U\cap V = \emptyset$. Then $X = (U\cap V)^{c} = U^{c} \cup V^{c}$.
Since $U$, $V$ are open, $U^{c} = V(I)$, $V^{c} = V(J)$ for some ideals $I,J$ of $k[x_{1},...,x_{n}]$. We have then that $X = V(I) \cup V(J) = V(IJ)$.
$I,J\neq (0)$ since $V(I),V(J)$ are proper subsets of $X$. So $IJ \neq (0)$ also since $k[x_{1},...,x_{n}]$ is an integral domain.
Thus there exists an $f\in IJ\setminus \{0\}$ such that $f$ vanishes at all points of $X$. Note that $f$ must in fact be nonconstant.
This is now where we use the fact that $k$ is infinite. Since $f$ is nonconstant, there exists an $i\in\{1,...,n\}$ such that $f$ has a nonzero term with a nonzero power of $x_{i}$ in it. Suppose for convenience of notation that $i=1$. We can then write $f = g_{m}(x_{2},...,x_{n})x_{1}^{m}+ ...+ g_{0}(x_{2},...,x_{n})$ for some $g_{0},...,g_{m}\in k[x_{2},...,x_{n}]$, $g_{m}\neq 0$, and $m>0$. So there is a $(a_{2},...,a_{n})\in k^{n-1}$ with $g_{m}(a_{2},...,a_{n})\neq 0$.
Then $f(x_{1},a_{2},...,a_{n})\in k[x_{1}]$ is nonzero, so can only have finitely many roots. This is a contradiction since $k$ is infinite, and $f$ must vanish at all points of $X=k^{n}$.
What's going on here is greatly clarified if you think of the two coordinate spaces of the product as being separate, totally unrelated spaces. So instead of thinking of $U$ and $V$ as both being subsets of the same polynomial ring $k[x]$, they are now subsets of two separate polynomial rings: $U\subseteq k[x_1]$ and $V\subseteq k[y_1]$. Now it's easy to guess what $Y$ should be: just take the union of $U$ and $V$ inside $k[x_1,y_1]$.
In terms of your original notation, this means $$Y=\{f(x_1):f(x)\in U\}\cup\{g(y_1):g(x)\in V\}.$$ I'll leave it to you to verify that this $Y$ works.
Best Answer
No, an arbitrary Zariski dense subset need not contain any non-empty Zariski open subset. Take $S = \mathbb{Z} = \{ \dots, -2, -1, 0, 1, 2, \dots \} \subset \mathbb{A}^1_{\mathbb{C}}$. Since the non-trivial Zariski closed subsets of $\mathbb{A}^1_{\mathbb{C}}$ are the finite sets, it is easy to check that the Zariski closure of $S$ is $\mathbb{A}^1_{\mathbb{C}}$ and that the only open set $U \subset S$ is $U = \emptyset$.
Edit. You probably want to prove the following stronger statement: If $U$ is Zariski dense in $\mathbb{A}^n_K$ and $f$ is any non-constant polynomial, then $D(f) \cap U$ is Zariski dense in $\mathbb{A}^n_K$. Indeed, suppose for contradiction that it were not. Then $D(f) \cap U \subseteq Z(g)$ for some polynomial $g$ (where $Z(g)$ is the zero locus of $g$). But then $$U = (D(f) \cap U) \cup (U \setminus D(f)) \subseteq Z(g) \cup Z(f) = Z(fg),$$ contradicting the assumed density of $U$.