= Solution
{c}
\b[Yes: every scheme is a Kolmogorov space.] Choose an <affine open subscheme> $V=\operatorname{Spec}A$ 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, choose
$$
a\in\mathfrak p_x\setminus\mathfrak p_y.
$$
The <principal open subscheme>
$$
\boxed{U=D(a)=\{\mathfrak p\in\operatorname{Spec}A:a\notin\mathfrak p\}}
$$
contains $y$ 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 $(*)$ in every case, proving <schemes are Kolmogorov spaces>.
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