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, observe that there are continuum many open sets, and then the Borel sets proceed in a hiearchy of length $\omega_1$, each time preserving the continuum number of sets at each level.
One can give very concrete examples. For example, the set of reals that code a binary relation on the natural numbers that is a well order is a complete $\Pi^1_1$ set, and cannot be Borel.