You can show that there are $\mathfrak{c} = 2^{\aleph_0}$ Borel subsets of the real line, and so by Cantor's Theorem ($|X| < | \mathcal{P} (X)|$) it follows that there are non-Borel subsets of $\mathbb{R}$.
To see that there are $\mathfrak{c}$-many Borel subsets of $\mathbb{R}$, we can proceed as follows:
- define $\Sigma_1^0$ to be the family of all open subsets of $\mathbb{R}$;
- for $0 < \alpha < \omega_1$ define $\Pi_\alpha^0$ to be the family of all complements of sets in $\Sigma_\alpha^0$ (so that $\Pi_1^0$ consists of all closed subsets of $\mathbb{R}$);
- for $1 < \alpha < \omega_1$ define $\Sigma_\alpha^0$ to be the family of all countable unions of sets in $\bigcup_{\xi < \alpha} \Pi_\xi^0$.
Then you can show that $B = \bigcup_{\alpha < \omega_1} \Sigma_\alpha^0 = \bigcup_{\alpha < \omega_1} \Pi_\alpha^0$ is the family of all Borel subsets of $\mathbb{R}$. Furthermore, transfinite induction will show that $| \Sigma_\alpha^0 | = \mathfrak{c}$ for all $\alpha < \omega_1$, which implies that $\mathfrak{c} \leq | B | \leq \aleph_1 \cdot \mathfrak{c} = \mathfrak{c}$.
Specific examples of non-Borel sets are in general difficult to describe. Perhaps the easiest to describe is a Vitali set, obtained by taking a representative from each equivalence class of the relation $x \sim y \Leftrightarrow x -y \in \mathbb{Q}$. Such a set is not Lebesgue measurable, and hence not Borel. Another example, due to Lusin, is given in Wikipedia.