TITLE: Problems where we can't make a canonical choice, solved by looking at all choices at once
QUESTION [45 upvotes]: It's a common theme in mathematics that, if there's no canonical choice (of basis, for example), then we shouldn't make a choice at all. This helps us focus on the heart of the matter without giving ourselves arbitrary stuff to drag around.
However, in this question, I'm looking for examples of problems solved by a specific type of "not making a choice" - namely, making all available choices, and looking at all the end results together as a whole. We can't necessarily discern any individual piece, but the average behavior, or some other information about the big picture, provides (or at least points towards) a solution.
I really wish I had an example of this phenomenon to provide, but even one escapes me at the moment, which is what spurred me to ask this question. I imagine combinatorics is full of examples; unfortunately I haven't really studied that field in any depth yet.
Something close to what I'm after is Burnside's (a.k.a. not-Burnside's) Lemma. There's no good way of directly counting orbits, i.e. choosing to look at a particular orbit one at a time, so we just look at the average number of fixed points of elements of $G$ (I'm reluctant to call this an example of the kind of result I'm looking for because I'm not entirely clear on why the fixed points of an element should be thought of as substitutes for orbits. Perhaps that's a separate question).
This question is in a similar vein.

REPLY [10 votes]: General topology as found in textbooks seems to be chock-full of examples where the axiom of choice seems to be (unconsciously?) invoked, and unnecessarily if one follows Eilenberg's advice to avoid subscripts! Andrej Bauer in his BAMS expository article on constructive mathematics, page 484, cites an example from Engelking's topology book. Here is an example from the famous textbook by Munkres (page 166 in the 1975 edition).

Every compact subset of a Hausdorff space is closed.
Proof. Let $Y$ be a compact subset of the Hausdorff space $X$. We shall prove that $X - Y$ is open, so that $Y$ is closed.
Let $x_0$ be a point of $X - Y$. For each point $y$ of $Y$, let us choose disjoint neighborhoods $U_y$ and $V_y$ of the points $x_0$ and $y$, respectively (using the Hausdorff condition). The collection $\{V_y|y \in Y\}$ is a covering of $Y$ (...)

Okay, stop right there. We can see where he's going with this, but do we have to make infinitely many choices like this? No, of course not. You could say instead

Let $x$ be a point of $X - Y$. The collection $\{V \subset X \;\text{open}| \exists U\; \text{open}: x \in U\; \text{and}\; U \cap V = \emptyset\}$ covers $Y$, because for each $y \in Y$ there are disjoint open neighborhoods $V$ of $y$ and $U$ of $x$, using the Hausdorff condition. (...)

If you are sensitive to the axiom of choice and look around, you find the same choice-y subscript-y device used all over the place, esp. in proofs of lemmas involving compactness arguments (such as the tube lemma).
What seems to be going on in such cases is that one (knowingly or not) has an entire or total relation $R$ from a set $A$ to a set $B$ (meaning that for all $a \in A$ there exists a $b \in B$ with $R(a, b)$), but one has resorted to the axiom of choice which guarantees the existence of a function $f \subseteq R$, typically denoted with a subscript: $a \mapsto b_a$. The advice is to look at all choices instead, i.e., work with the entire relation $R$ directly, and dispense with choice functions altogether.
Or, don't. It's your choice. But if some astute student asks whether you secretly used AC there and is that actually necessary? then you'll be glad if you had the other argument ready.