Some comments, unrelated to the other answer, on what happens in the case of the Zariski topology. Here the extra axiom is

$$V(I(S_1 \cup S_2)) = V(I(S_1)) \cup V(I(S_2))$$

(where, to fix notation, the $S_i$ are subsets of $k^n$, $I$ sends a subset of $k^n$ to the ideal of functions vanishing on it in $k[x_1, ..., x_n]$, and $V$ sends a subset of $k[x_1, ..., x_n]$ to its zero locus in $k^n$.) Now, we have $I(S_1 \cup S_2) = I(S_1) \cap I(S_2)$ for abstract reasons (the adjointness implies that $V$ and $I$ both send colimits to limits, or equivalently suprema to infima), but we don't have $V(T_1 \cap T_2) = V(T_1) \cup V(T_2)$ for arbitrary subsets $T_1, T_2$ of $k[x_1, ..., x_n]$ for rather silly reasons, e.g. $T_1$ and $T_2$ could be disjoint but generate the same ideal. Hence the condition mentioned in Nacho Lopez's answer doesn't apply here.

At best we can hope to have $V(T_1 \cap T_2) = V(T_1) \cup V(T_2)$ for $T_1, T_2$ ideals, which would still be enough to get the extra axiom, but note that this fails when $k$ has zero divisors, e.g. if $k = \mathbb{Q}[a, b]/(ab)$ then we could take $T_1 = (ax_1), T_2 = (bx_2)$, and then $T_1 \cap T_2 = (0)$ has zero locus $k^n$ but $V(T_1) \cup V(T_2)$ is much smaller, e.g. it doesn't contain $(1, 1)$. That strongly suggests it can't be true for abstract reasons: at the very least we need to use somewhere the fact that $k$ is an integral domain.

What is true for abstract reasons is that $T_1 \cap T_2 \subseteq T_1, T_2$, hence $V(T_1), V(T_2) \subseteq V(T_1 \cap T_2)$, hence $V(T_1) \cup V(T_2) \subseteq V(T_1 \cap T_2)$. In particular, we get the inclusion

$$V(I(S_1)) \cup V(I(S_2)) \subseteq V(I(S_1) \cap I(S_2)) = V(I(S_1 \cup S_2))$$

for abstract reasons. But to get the other inclusion we need to actually do something. For example:

Suppose $v \in V(I(S_1 \cup S_2))$, so $f(v) = 0$ for all $f \in I(S_1 \cup S_2)$. In particular $g(v) h(v) = 0$ for all $g \in I(S_1), h \in I(S_2)$ (since $IJ \subseteq I \cap J$ for ideals). If $g(v) = 0$ for all $g \in I(S_1)$ then $v \in V(I(S_1))$; otherwise, there is some $g$ such that $g(v) \neq 0$, hence from $g(v) h(v) = 0$ for all $h$ we deduce that $h(v) = 0$ for all $h$, and then $v \in V(I(S_2))$.

So, abstractly, here is what we did:

• In the domain of $V$ and the range of $I$ we restricted our attention to ideals.
• On ideals we introduced a new operation, namely ideal product, satisfying $I(S_1) I(S_2) \subseteq I(S_1 \cup S_2)$.
• We showed that $V(IJ) \subseteq V(I) \cup V(J)$. (For this step we needed to use the fact that $k$ is an integral domain.)

This is enough: once these conditions are met we then have

$$V(I(S_1 \cup S_2)) \subseteq V(I(S_1) I(S_2)) \subseteq V(I(S_1)) \cup V(I(S_2)).$$

To formalize this argument, we can think of ideal product as a monoidal operation on the poset of ideals making it a monoidal category. The poset of subsets of $k^n$ also naturally has a monoidal operation given by union. The condition on $I$ above is that $I$ is a lax monoidal functor with respect to these monoidal operations, and the condition on $V$ above is that $V$ is an oplax monoidal functor with respect to these monoidal operations. All told, we get the following:

Let $F : P \to Q$ and $G : Q \to P$ be an antitone Galois connection between posets $P, Q$. Suppose furthermore that $P$ has finite coproducts and that $Q$ is equipped with a monoidal operation $\otimes$ with respect to which $F$ is lax monoidal (where $P$ is equipped with the coproduct) and $G$ is oplax monoidal. Then the closure operator $GF$ on $P$ preserves binary coproducts.

But this is somehow unsatisfying.