$\operatorname{Spec} (A)$ as a topological space satisfying the $T_0$ axiom

(expanding my comment to a (partial) answer)

Use that $X$ is $T_0$ iff for all $x,y \in X$, if $x \in \overline{\{y\}}$ and $y \in \overline{\{x\}}$ then $x = y$.

Proof: from left to right: If $x \neq y$, the $T_0$ gives us an open set such that $x \in U, y \notin U$, say (the other case is similar). But then $x \in \overline{\{y\}}$ implies $y \in U$, as $U$ is a neighbourhood of $x$, so intersects $\{y\}$. Contradiction.

From right to left: if $x \neq y$, then $x \notin \overline{\{y\}}$ or $y \notin \overline{\{x\}}$, say the former. Then $U = X \setminus \overline{\{y\}}$ is open, contains $x$, but not $y$ (in the other case, of course, the other way around).

You already know $\overline{\{x\}} = V(\mathfrak{p}_x)$, so you need to prove $$V(\mathfrak{p}_x) = V(\mathfrak{p}_y) \rightarrow x=y$$

which is something that you might know how to do (I don't as I don't have this book).

Let $f$ in $A$. It defines a principal open set $U_f$. Namely : $$ U_f = \{ x\in X \mid f\not\in \mathfrak p_x \}$$

Mouse over the grey box only if you want more than the above hint.

If you have two different points of $X$, you can certainly find a $f$ in $A$ which is in one and only one of the two ideals $\mathfrak p_x$ and $\mathfrak p_y$. Then $U_f$ contains exactly one of both points $x$ and $y$.