 Solution

Yes: every scheme is a Kolmogorov space. Choose an affine open subscheme $V=\mathrm{Spec}A$ containing $x$. If $y\notin V$, take $U=V$.

Otherwise, the distinct points $x,y\in V$ correspond to distinct prime ideals ${\mathfrak{p}}_x,{\mathfrak{p}}_y\subseteq A$. There is an element in one of these prime ideals and not the other. After interchanging the names of the two points if necessary, chooseThe principal open subschemecontains $y$ and does not contain $x$. It is an open subset of $V$, hence an open subset of $X$ because $V$ is open in $X$. This establishes $(\ast )$ in every case, proving schemes are Kolmogorov spaces.