This is a very standard result that can be found in almost all textbooks which provide an introduction to (un)countability. The most common proof uses Cantor's diagonal argument, which is echoed below.
Suppose $\mathbb{R}$ is countable. Then we can enumerate its elements as follows:
$$\mathbb{R} = \{ x_1, x_2, x_3, \cdots \}$$
(If $f : \mathbb{R} \to \mathbb{N}$ is a bijection then we can take $x_i = f^{-1}(i)$.)
Each $x_i$ has a unique decimal expansion in which infinite strings of $0$s are taken if there is any ambiguity. So write
$$\begin{align}x_1 &= n_1 + 0.d_{11}d_{12}d_{13} \cdots \\ x_2 &= n_2 + 0.d_{21}d_{22}d_{23} \cdots \\ x_3 &= n_3 + 0.d_{31}d_{32}d_{33} \cdots \\ &\ \vdots\end{align}$$
where $n_i \in \mathbb{Z}$ for each $i$ and $d_{ij} \in \{ 0, 1, \cdots, 9 \}$.
Construct $x \in \mathbb{R}$ as follows. Let
$$x = 0.d_1d_2d_3\cdots$$
where $d_i = 0$ if $d_{ii} \ne 0$ and $d_i = 1$ if $d_{ii}=0$. Then in particular $d_i \ne d_{ii}$ for any $i$. Since $x \in \mathbb{R}$ and $\mathbb{R}$ is supposed to be countable, we must have $x=x_i$ for some $i$, but that would mean that $d_i = d_{ii}$... contradiction!
