Let $V(\mathfrak{a})$ be all ideals in ${\rm Spec}(A)$ that contain ideal $\mathfrak{a}$. Then $V(\mathfrak{a} \cap \mathfrak{b}) = V(\mathfrak{a}) \cup V(\mathfrak{b})$.
$\mathfrak{p} \in$ RHS means $\mathfrak{p}$ contains one of the two so obviously contains the intersection. OTOH, $\mathfrak{p} \in$ LHS, then $\mathfrak{p} \supset \mathfrak{a} \cap \mathfrak{b}$. For all $a \in \mathfrak{a}$. Let $a \in \mathfrak{a}, b \in \mathfrak{b}$, then $ab \in \mathfrak{a} \cap \mathfrak{b}$ since each absorbs the whole ring $A$. Then we have that either $a$ or $b \in \mathfrak{p}$. What now?
We can show the following equalities $$V(\mathfrak{a} \mathfrak{b}) = V(\mathfrak{a} \cap \mathfrak{b}) = V(\mathfrak{a}) \cup V(\mathfrak{b}).$$
Because $\mathfrak{a}\mathfrak{b} \subseteq \mathfrak{a}\cap \mathfrak{b}$, we have $$V(\mathfrak{a}\cap \mathfrak{b}) \subseteq V(\mathfrak{a}\mathfrak{b})$$ because any prime that contains $\mathfrak{a} \cap \mathfrak{b}$ must also contain $\mathfrak{a}\mathfrak{b}$.
Then, we have $V(\mathfrak{a}) \cup V(\mathfrak{b}) \subseteq V(\mathfrak{a} \cap \mathfrak{b}$ by what you've already shown. So, we're left with one containment to show, $V(\mathfrak{ab}) \subseteq V(\mathfrak{a}) \cup V(\mathfrak{b}).$
To show this, suppose a prime $\mathfrak{p}$ contains $\mathfrak{a}\mathfrak{b}$ and suppose for contradiction that it contains neither $\mathfrak{a}$ or $\mathfrak{b}$. Then, there must be elements $a \in \mathfrak{a} \backslash \mathfrak{p}, b \in \mathfrak{b} \backslash \mathfrak{p}.$ Because $\mathfrak{p}$ is prime, this implies that $ab \in \mathfrak{a}\mathfrak{b} \backslash \mathfrak{p}$ (as $ab \in \mathfrak{p} \Rightarrow a \in \mathfrak{p} \text{ or }b \in \mathfrak{p}.$)
But this is a contradiction because $\mathfrak{p}$ contains $\mathfrak{a}\mathfrak{b}$. Hence, $\mathfrak{p}$ must contain one of $\mathfrak{a}, \mathfrak{b}$, finishing the proof.