The Borel measures on $\Omega$ can be identified with continuous linear functionals $\newcommand{\bR}{\mathbb{R}}$ $\mu:C(\Omega)\to\bR$. Assume that $\Omega$ is compact so $C(\Omega)$ is Banach space. Denote by $\Vert-\Vert$ the sup norm on $C(\Omega)$ and by $\newcommand{\eM}{\mathscr{M}}$ $\eM(\Omega)$ the dual of $C(\Omega)$ equipped with the dual norm
$$
\Vert \mu\Vert_*:=\sup_{\Vert f\Vert\leq 1}|\mu(f)|.
$$
Set
$$
C(\Omega)_+:=\big\{ f\in C(\Omega):\;\; f(x)\geq 0,\;\;\forall x\in\Omega)\big\}.
$$Let $\mu\in \eM(\Omega)$ and $f\in C(\Omega)$. Then, for any $f\in C(\Omega)_+$ we have (see Theorem 4.3.2. of *R.E. Edwards: Functional Analysis*)
$$
\mu_+(f)=\sup_{0\leq g\leq f} \mu(g).
$$
Let $\mu,\nu\in \eM(\Omega)$, and $f\in C(\Omega)_+$. Then for any $0\leq g\leq f$ we have

$$
\mu(g)-\nu(g)\leq \Vert\mu-\nu\Vert_*\Vert g\Vert \leq \Vert\mu-\nu\Vert_*\Vert f\Vert.
$$
Hence, for any $0\leq g\leq f$
$$
\mu(g)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert+\nu(g),
$$
and, symmetrically,
$$
\nu(g)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert+\mu(g).
$$
Taking the sup on both sides of the above inequalities we deduce
$$
\mu_+(f)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert+\nu_+(f)\implies \mu_+(f)- \nu_+(f)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert,
$$
$$
\nu_+(f)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert+\mu_+(f)\implies \nu_+(f)- \mu_+(f)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert.
$$
Hence
$$
\big\vert (\mu_+-\nu_+)f\big\vert\leq \Vert\mu-\nu\Vert_*\Vert f\Vert.
$$
This implies
$$
\Vert \mu_+-\nu_+\Vert_*\leq \Vert\mu-\nu\Vert_*.
$$