Suppose $\{x\}$ is an atom of $|Du|$. A simple way to arrive at a contradiction is to show that atoms of $Du$ (and hence $|Du|$) correspond to jump discontinuities of $u$, since jump discontinuities are never Lebesgue points. This should make sense, since the distributional derivative of the Heaviside function is a Dirac delta.
To see this, first note that $\mu:=Du$ is the Lebesgue-Stieljes measure associated to $u$, since for any compactly supported smooth $v$, by integration by parts for Lebesgue-Stieljes measures,
$$\int v \ d\mu = -\int u \ dv = -\int uv’ \ dx$$
Using continuity of measure, the definition of Lebesgue-Stieljes measure, and the fact that BV functions have left and right limits, we see that
\begin{align*} mu(\{x\}) &= \lim_{r\to 0}\mu((x-r,x+r]) \\ &= \lim_{r\to 0} (u(x+r) - u(x-r))\\ &= u(x^+) - u(x^-) > 0 \end{align*}
Therefore, $u$ has a jump at $x$.