For real numbers $z$ and $w$, $|(1+z)(1+w)-1| \leq (1+|z|)(1+|w|)-1$.

To answer the solution-verification part of the question, this step is wrong.

one obtains

\begin{align} (1+z)|(1+w)| \leq |(1+z)||(1+w)| \end{align}

Since $(1+w) \leq |(1+w)|$, it follows that

\begin{align} (1+z)(1+w) \leq |(1+z)||(1+w)| \end{align}

The above is of the form $\, a \cdot |b| \le c \implies a \cdot b \le c\,$, which does not hold true in general. For example, if $\,a=b=-2, \,c=1\,$ then $\,a \cdot |b| = -4 \le 1 = c\,$, but $\,a \cdot b = 4 \gt 1 = c\,$. The implication does hold true for $\,a \ge 0\,$, but here $\,a = 1 + z\,$ which is not necessarily positive.

The correct way to derive the inequality is to use that $\,|a \cdot b| = |a| \cdot |b|\,$, then:

$$ (1+z)\cdot (1+w) \;\leq\; |(1+z)\cdot(1+w)| \;=\; |1+z|\cdot|1+w| $$

We have :

$(1+z)(1+w)-1 = z+w+zw$

Thus :

$|(1+z)(1+w)-1| = |z+w+zw| \le |z| + |w| + |zw| = (1+|z|)(1+|w|) -1$

QED