When does a Galois connection induce a topology? Let $(X,\leq)$ and $(Y,\leq)$ by partially ordered sets.  Recall that a(n antitone)  Galois connection between $X$ and $Y$ is a pair of order-reversing maps 
$\Phi: X \rightarrow Y, \ \Psi: Y \rightarrow X$
such that for all $x \in X,\  y \in Y$, $x \leq \Psi(y) \iff \Phi(x) \geq y$.  
In this situation, the composite $\Psi \circ \Phi$ (resp. $\Phi \circ \Psi$) is a closure operator $\operatorname{cl}$ on $X$ (resp. $Y$): that is,
(C1) For all $x \in X$, $x \leq \operatorname{cl}(x)$.
(C2) For all $x_1,x_2 \in X$, $x_1 \leq x_2 \implies \operatorname{cl}(x_1) \leq \operatorname{cl}(x_2)$.
(C3) For all $x \in X$, $\operatorname{cl}(\operatorname{cl}(x)) = \operatorname{cl}(x)$.  
Let us consider the special case where $X$ is the power set of some set $\mathbb{X}$, partially ordered by inclusion.  In order for the above closure operator to define a topology on $\mathbb{X}$ -- i.e., be a "Kuratowski closure operator" -- we need also
(C4) For all $x,y \in X$, $\operatorname{cl}(x \cup y) = \operatorname{cl}(x) \cup \operatorname{cl}(y)$.  
It is easy to see that (C4) is not automatic for the closure operator associated to a Galois connection -- in fact, every closure operator is induced by some Galois connection,
and there are lots of closure operators which do not satisfy (C4).  
However, in practice, it seems quite often to be the case that at least one of the two closure operators induced by a Galois connection is a topological closure operator.  Examples:
(1) The Krull topology on the automorphism group of an infinite Galois extension (coming from the usual Galois correspondence).
(2) For a field $K$, the Zariski topology on $K^n$ (coming from the Galois connection arising in the Nullstellensatz).
(3) For a field $K$, the Harrison topology on the real spectrum of $K$ (coming from the Galois connection induced by the "incidence relation" $x \in P$ for $x$ an element of $K$ and $P$ an ordering on $K$).

Question 1: Is there some natural abstract condition on the Galois connection one can impose in order to ensure that one of the two closure operators is a Kuratowski closure operator?

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Question 2: Are there are other examples of topologies arising from Galois connections in an interesting way?  [As above, every topology arises from some Galois connection, yet in a tautological way.]

 A: All of the Galois connections I know involving a power set arise from a relation $R : X \times Y \to 2$ as described for example here. As you observe, this relation can often be used to define a topological space. However, it can always be used to define a Chu space, which is in a precise sense a kind of generalized topological space. 
Some definitions. The category $\text{Chu}$ of Chu spaces (in $\text{Set}$, over $2$) has as objects triplets $(X, Y, R)$ where $X, Y$ are sets and $R$ is a function $X \times Y \to 2$. A morphism $(X_1, Y_1, R_1) \to (X_2, Y_2, R_2)$ of Chu spaces is an adjunction: that is, it is a pair of functions $f_{\ast} : X_1 \to X_2$ and $f^{\ast} : Y_2 \to Y_1$ such that
$$R_2(f_{\ast}(x_1), y_2) = R_1(x_1, f^{\ast}(y_2)).$$
Many familiar categories turn out to be full subcategories of $\text{Chu}$. 
Example. Suppose the map $Y \ni y \mapsto \{ x : xRy \forall y \in Y \} \in 2^X$ is injective. The full subcategory of Chu spaces with this property may be regarded as sets together with a distinguished collection of subsets, where $R$ is the incidence relation "the point $x \in X$ lies in the subset $S \in Y$," and where $f^{\ast}$ is consequently always the preimage along $f_{\ast}$, so the morphisms are precisely maps of sets such that the preimage of a distinguished subset is distinguished.
Subexample. In particular, topological spaces are such Chu spaces, and the notion of morphism agrees. Hence $\text{Top}$ is a full subcategory of Chu spaces. 
This is the precise sense in which Chu spaces generalize topological spaces: being a topological space is a property of a Chu space, in the same way that being an abelian group is a property of a group. 
Subexample. Similarly, the category $\text{Meas}$ of measurable spaces is a full subcategory of Chu spaces. 
Example. Let $V$ be a vector space over $\mathbb{F}_2$. Then the dual pairing $\text{eval} : V \times V^{\ast} \to \mathbb{F}_2$ defines a Chu space. This inclusion is full.
Example. Similarly, let $B$ be a Boolean algebra with Stone space $S(B)$. Then the evaluation map $\text{eval} : B \times S(B) \to 2$ defines a Chu space. This inclusion is full. 
Example. Let $P$ be a poset and let $2^P$ denote the poset of order-preserving functions $P \to 2$, where $2 = \{ 0, 1 \}$ and $0 < 1$. The evaluation map $\text{eval} : P \times 2^P \to 2$ defines a Chu space. This inclusion is full.
Example. $\text{Chu}$ is equipped with a natural involution switching $X$ and $Y$. Applying this involution to any of the above examples gives other examples. When applied to various subcategories of $\text{Chu}$ this involution reproduces many familiar dualities. For example, when applied to Boolean algebras, the involution reproduces Stone duality. 
I haven't played much with Chu spaces, but my intuition is that one can think of $X$ as a set of "states" and $Y$ as a set of "observations" that one can make about states, with $R$ confirming whether or not a given observation is true of a given state. But then one can switch the roles of states and observations! 
A: 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. 
A: I'll try to give an approximation from Category Theory. First, a partially ordered set $(X,\leq)$ can be regarded as a category with objects the elements of $X$ and one arrow $x\to x'$ iff $x\leq x'$. From this point of view, an order-preserving function between partially ordered sets is just a functor between the respective categories. Your Galois connection $\Phi,\Psi$ is a pair of functors $\Phi:X\to Y^{op}$, $\Psi:Y^{op}\to X$ and $\Phi$ is a left adjoint of $\Psi$. 
Now to closure operators. A closure operator on $T$ on $X$ is just an idempotent monad on the category $X$. The fact that any closure operator comes from a Galois connection is a particular case of the general fact that any monad on a category is induced by an adjunction (usually there are several adjunctions that induce the same monad). 
The condition $\operatorname{cl}(x \cup y) = \operatorname{cl}(x) \cup \operatorname{cl}(y)$ is saying that your idempotent monad $\operatorname{cl}$ on $X$ preserves binary coproducts. What you write as union would be the supremum of two elements $(x\vee x)$ in your partially ordered set, which is just the coproduct in the corresponding category. 
A standard fact about adjunctions is that the left adjoint (in your case $\Phi$) always preserves colimits, and in particular coproducts. Then, one condition that ensures that $\operatorname{cl}$ preserves joins is to require $Y^{op}$ to have and the right adjoint $\Phi:Y^{op}\to X$ to preserve binary coproducts. $Y^{op}$ has binary coproducts when 
when $(Y,\leq)$ has meets of two elements $(y\wedge y')$, and $\Psi$ preserves them when $\Psi(y\wedge y')=\Psi(y)\vee\Psi(y')$. 
Hope this answers to some extent your Question 1. 
