Nice Example of a Non-Borel Set in $\mathbb{R}^2$? Can someone explain, intuitively, what the meaning of a non-Borel set in  $\mathbb{R}^2$ is? Is it some sort of collection of points with some weird property that it cannot be formed from countable unions or intersections of the open sets of the Euclidean topology on the 2D plane? What does this look like?
Though there are explanations, is there some nice way of understanding this?
 A: To start off with, let me confirm that your guess is correct: a non-Borel subset of $\mathbb{R}^2$ is a set of points in the plane which can't be "built up" from open sets via the standard operations (= countable unions, countable intersections, complements).
Now in practice, "most" reasonably-definable sets of points in the plane are Borel, so it's a bit hard to give an example of a non-Borel set. To do this, in my opinion it's easier to work in Cantor space $2^\omega$ rather than $\mathbb{R}^2$, and then move the example we get to $\mathbb{R}^2$.
Specifically, note that we can represent binary relations on the naturals by subsets of the naturals: namely, given a binary relation on naturals, $R\subseteq\mathbb{N}^2$, we get a corresponding set $$Set(R)=\{\langle a,b\rangle: aRb\}$$ (where $\langle\cdot,\cdot\rangle$ is your favorite pairing operation on naturals).
Of course we can extend this well beyond binary relations, but let's stick to them for now.
Now given a set of binary relations $\mathbb{K}$, we get a corresponding set of sets of natural numbers: $$Set(\mathbb{K})=\{Set(R): R\in\mathbb{K}\}.$$ Thinking of $Set(\mathbb{K})$ as a subset of Cantor space, we can ask about its topological complexity. For example, the set of points in Cantor space corresponding to transitive relations is closed but not open.
So we can phrase the search for natural non-Borel sets as:

Is there a natural class of relations $\mathbb{K}$ such that $Set(\mathbb{K})$ is not Borel?

The answer is yes. Say that a binary relation $R\subseteq\mathbb{N}^2$ is well-founded if there is no infinite sequence $(a_i)_{i\in\mathbb{N}}$ of naturals such that $$a_0Ra_1Ra_2R...$$ For example, if $R$ is the usual ordering on the naturals, then $R$ is well-founded: this is exactly what makes proof by induction work. Conversely, a well-founded relation can never be reflexive: if $bRb$, just let $a_i=b$ ...
It turns out that the class $\mathbb{W}$ of well-founded binary relations on $\mathbb{N}$ yields a non-Borel set. Any "reasonable" injection from Cantor space to $\mathbb{R}$ - say, the usual identification of the Cantor space and the Cantor set - then gives us a corresponding non-Borel subset of $\mathbb{R}$.

Let me end by mentioning two further points.

*

*We may reasonably ask, "What do sets look like past the Borel sets?" This takes us into the subject of descriptive set theory. The Borel sets correspond, it turns out, to the very lowest level of the projective sets, while our non-Borel example above lives "one level up" (precisely: $Set(\mathbb{W})$ is $\Pi^1_1$-complete, also called "properly coanalytic"). In general you may be interested in the Wadge hierarchy.


*The other issue is the role of the axiom of choice. Asaf commented that the axiom of choice is necessary to show that non-Borel sets exist in the first place; that is, it's consistent with ZF (= set theory without choice) that every set of reals is Borel. But the non-Borel set I exhibited above was pretty simple to describe, and didn't seem to need choice; moreover, the proof that it's non-Borel (which I admittedly didn't give) is quite direct. So, what gives?
It turns out that the issue here is the definition of "Borel." There are two definitions of "Borel set." The first is the usual one: $X$ is Borel if $X$ is an element of every $\sigma$-algebra of sets containing the open sets. This is the "top-down" definition of Borel. The second is more explicit: a set $X$ is Borel if it has a Borel code, that is, an actual process for building it from open sets via countable unions, countable intersections, and complements. This is a "bottom-up" definition of Borel-ness. (See e.g. here.)
The point is that the equivalence between these notions is not provable in ZF alone! ZF proves that every set which is Borel in the second sense is Borel in the first sense, but not conversely. In fact, ZF proves that $Set(\mathbb{W})$ is not Borel in the second sense, but ZF can't prove the existence of a set which is not Borel in the first sense. (For this reason, when we're working in set theory without choice, "Borel" is generally taken to refer to the second notion above.)

EDIT: Very belatedly, let me observe that from the point of view of Borel (as opposed to open) sets, there's no real difference between $\mathbb{R}^2$ and $\mathbb{R}$; precisely, there is a bijection $b:\mathbb{R}^2\rightarrow\mathbb{R}$ which induces an isomorphism between the algebras of Borel sets of $\mathbb{R}^2$ and $\mathbb{R}$. Similarly, there's no real difference between any of the finite powers of $\mathbb{R}$, or even stranger things like the powerset of the natural numbers, $\mathcal{P}(\mathbb{N})$ (topologized by thinking about it as the $\mathbb{N}$-fold product of the discrete two-point space). So examples of non-Borel sets in any of these variants can be "translated" to $\mathbb{R}^2$ via $b$.
In the particular case of $2^\mathbb{N}$, I just learned about a striking example of a very non-Borel (= $\Sigma^1_1$-complete) set with a very elementary definition indeed: the set of difference sets! Given $A\subseteq\mathbb{N}$, let $$\mathscr{D}(A)=\{\vert x-y\vert: x,y\in A\}.$$ The range of the operator $\mathscr{D}$ is trivially $\Sigma^1_1$, and Schmerl showed that that's sharp. I would tentatively say that this is the simplest non-Borel, let alone $\Sigma^1_1$-complete, set in any appropriately "Euclidean-like" (= Polish) space I can think of.
