Can we get un-obvious results by defining sophisticated topologies?

What you've written sounds like a very good description of the entire field of functional analysis. One constructs topologies on various spaces of functions and uses topological properties to prove things.

For instance, if one is looking for a function with certain properties (say, one that solves a certain differential equation), a common technique is to construct a sequence of functions that "approximately" have those properties, and use compactness to claim the sequence has a limit point. Then, using some sort of continuity, claim that the limit point has exactly the desired properties. Often the topology has been specifically constructed to provide such continuity.

Examples include $L^p$ spaces, Sobolev spaces, spaces of continuous functions and of distributions (generalized functions), and many more.

Let $G$ be an infinite graph, such that every finite subgraph of $G$ can be colored with $k$ colors. Is it necessarily the case that $G$ can be colored with $k$ colors?

The answer is yes, and to prove this we will begin by defining a topology on the space of all colorings of $G$.

Let $V$ be the set of vertices of $G$. A coloring of $G$ - not necessarily legal - is just a function $f : V \to A$, where $A$ is the finite set of $k$ colors. Let us denote the space of all colorings by $X$.

We can topologize $X$ by taking a basic open set to be of the form $U_{v,c}$ where $v$ is a vertex of $G$, $c$ is some color in $A$, and $U_{v,c}$ is the set of those colorings $f \in X$ such that $f(v)=c$. A little thought reveals that this is just the space product space $A^V$ where $A$ is given the discrete topology.

There is a highly nontrivial theorem of Tychonoff that says that the product of compact spaces is again compact. So the space $X$ is compact. This implies that every collection of closed subsets of $X$ with the finite intersection property (the intersection of finitely many sets in the collection is nonempty) has a nonempty intersection.

Let $H$ be a finite subgraph of $G$. If $f$ is a coloring of $G$, we can obtain a coloring of $H$ by restricting $f$ to the set of vertices of $H$. Of course, if $f$ were a valid coloring of $G$, then restricting it to $H$ gives a valid coloring of $H$ - in $H$, there are less vertices and edges to worry about.

Conversely, let $f$ be a coloring of $G$, and suppose that restricting $f$ to each finite subgraph of $G$ gives a valid coloring. I claim that $f$ is a valid coloring of $G$, too. Suppose not. Then there are vertices $u,v$ connected by an edge such that $f(u) = f(v)$. But then the restriction of $f$ to the finite subgraph $H$, which consists of the vertices $u,v$ and the edge between them, is not a legal coloring. This contradicts our initial asumption.

So we are going to find a coloring of $f$ of $G$ whose restriction to each finite subgraph is legal and this will prove our theorem.

If $H$ is a finite subgraph of $G$, let $C_H$ denote the set of those colorings of $G$ whose restriction to $H$ is legal. By hypothesis $C_H$ is nonempty for every such $H$. Furthermore, the set $C_H$ is closed. For if $f$ is a coloring of $G$ which lies outside $C_H$, then there are vertices $u,v$ of $H$ connected by an edge in $H$ such that $f(u) = f(v)$. The set of $U$ of those $g$ which agree with $f$ on $u,v$ is open and it contains $f$, but it is disjoint from $C_H$ (no such $g$ can be valid when restricted to $H$). It follows that $C_H$ is closed, since its complement contains a neighborhood of each of its points.

Finally, the family ${C_H}$ enjoys the finite intersection property: if $H_1, ..., H_n$ are finite subgraphs, then a valid coloring of their union is a valid coloring of each of them individually. So $C_{H_1 \cup ... \cup H_n}$ is a nonempty subset of $C_{H_1} \cap ... C_{H_n}$.

By compactness the intersection of all the $C_H$, for $H$ a finite subgraph, is nonempty. If $f$ is an element of the intersection, then it is a valid coloring of each finite subgraph by definition, and hence, by the above, a valid coloring of $G$.

The above proof is very similar to the proof that the compactness theorem in topology yields the compactness theorem for propositional calculus; thus logical and topological compactness are distinct, but related notions.

Here's a great example of what you're looking for: Furstenberg's proof of the infinitude of primes. http://en.wikipedia.org/wiki/Furstenberg%27s_proof_of_the_infinitude_of_primes