If you assume the countable axiom of choice, then most sets
of reals are not Borel. Under AC, what you get is that
there are continuum many Borel sets, that is,
$2^{\aleph_0}$ many, but $2^{2^{\aleph_0}}$ many sets of
reals, so most sets of reals are not Borel. Under countable
choice $AC_\omega$, what we have is a surjection from the
reals onto the Borel sets, and this is still enough to
conclude that 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 hierarchy 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. 

Under $AC_\omega$, the collection of sets of reals
coded by such Borel codes is a $\sigma$-algebra containing
all open sets, and the smallest such, so it is exactly the
collection of Borel sets. The proof that it is a
$\sigma$-algebra uses $AC_\omega$, becasue when we have a countable
family of Borel sets, to make the code for their union, we
can simply glue together Borel codes for each of them,
provided we can choose representing Borel codes. Thus,
under $AC_\omega$, we can map the reals onto the set of
Borel sets. But by Cantor's theorem, we cannot map the
reals onto the power set of the reals, and so in this
sense, most sets of reals are not Borel.

Under $AC_\omega$, one can also give 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 (under $AC_\omega$) 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, under $AC_\omega$, since if it were, I could flip
the values on the diagonal and produce an analytic set that
is not a slice.