Proof, [Proposition 4.1.5, AP]. Let $I \subset M$ be a non-zero two-sided ideal. By Proposition 43.7.4, $I$ admits a non-zero projection $e \in I$. Let $\mathcal{P}\subset \text{Proj}(M)$ be a maximal family of mutually orthogonal projections equivalent to $e$. Since $\tau$ is faithful, $\tau(p) > 0$ for all $p \in \mathcal{P}$, so $\mathcal{P}$ is finite.

Let $q = 1_{M} - \sum_{p \in \mathcal{P}}p$, then by the comparability theorem, either $q \preceq e$ or $e \prec q$. If $e \prec q$, then there exists a subprojection $p \le q$ with $e \sim p$, which contradicts the maximality of $\mathcal{P}$. Thus $q \preceq e$.

Let $\mathcal{Q}= \mathcal{P}\sqcup \bracsn{q}$, then $p \preceq e$ for all $p \in \mathcal{Q}$ and $1_{M} = \sum_{p \in \mathcal{Q}}p$. For each $p \in \mathcal{Q}$, let $v_{p} \in M$ be a partial isometry such that $v_{p}v_{p}^{*} = p$ and $v_{p}^{*}v_{p} \le e$. As such, $p = v_{p}ev_{p}^{*} \in I$ for all $p \in \mathcal{Q}$. Therefore $1_{M} = \sum_{p \in \mathcal{Q}}p \in I$, and $I = M$.$\square$