It is not difficult to see that there are only continuum many Borel sets, that is, $2^{\aleph_0}$ many. But there are $2^{2^{\aleph_0}}$ many sets of reals. So most sets of reals are not Borel.
To see that there are only continuum many Borel sets, one can think about how the Borel sets are constructed. We begin with the basic open sets, of which there are countably many, and then systematically close under countable unions, intersections and complements. It follows that the Borel sets are constructed in a hiearchy of length $\omega_1$, and that every Borel set has a construction template, known as a Borel code, that details exactly how it was constructed from the basic open sets. One can think of the Borel code as a well-founded countable tree, whose leaves are labeled with basic open sets and whose other nodes are labeled with union, intersection and complement, meaning that this is the operation to be applied to the children node in order to know which set is coded at the parent node. Every such tree is a countable object, coded by a real. Thus, we can map the reals onto the set of Borel sets, and conversely, so they have the same size, without using AC.
One can also give very concrete examples of sets that are not Borel. For example, the set of reals known as WO is the set of reals that code a binary relation on the natural numbers that is a well order. This is a complete $\Pi^1_1$ set, and cannot be Borel.
To explain a bit more: if I have a relation $R$ on the natural numbers for which $\langle\mathbb{N},R\rangle$ is a well order, then I can code this relation $R$ as a single binary sequence, by placing a $1$ in the $2^n3^m$ digit exactly if $nRm$. The set of such sequences coding such well-orders has complexity $\Pi^1_1$, since to be a well-order means that it is a linear order (which is easy to express using only natural number quantifiers) plus the assertion that every subset has a minimal element (which is the universal quantifier over reals). But more, it is complete $\Pi^1_1$, which means that every $\Pi^1_1$ set reduces to WO. But the Borel sets are exactly the $\Delta^1_1$ sets, which are the sets that are both $\Pi^1_1$ and $\Sigma^1_1$, meaning that their complement is $\Pi^1_1$.
Basically, any set that is defined using well-foundedness will essentially involve $\Pi^1_1$ and WO, and take you out of the Borel context. The set of all well-founded countable trees, the set of well-founded countable relations, the set of well-orders, and so on are all complete $\Pi^1_1$ sets, and therefore not Borel.
One can similarly work on the analytic side, to come up with $\Sigma^1_1$ examples. There is a universal $\Sigma^1_1$ set, an analytic subset of the plane whose slices are all analytic sets, and such a set cannot be Borel, since if it were, I could flip the values on the diagonal and produce an analytic set that is not a slice.