A follow up for this question. If $\mathcal{A}$ is a unital von Neumann algebra and $a\in\mathcal{A}$ is a positive element such that $\|a\|=1$. Let $p \in\mathcal{A}$ be a non-zero projection. When does the following statement fail? Is there a counter-example?
$p\leq a$ iff, for every nozero projection $\mathcal{A}\ni q\leq p$, we have $\|a- (1-q)\|=1$.
Also, is there a short proof for when it's true?
$\implies)$ since from $p\leq a$ we know that $p=ap$. Then $ap=pa$ and $$ a=ap+a(1-p)=p+a(1-p).$$ If $q\leq p$ then $1-q=(p-q)+1-p$ and then $$ a-(1-q)=q+(a-1)(1-p). $$ Then $$ \|a-(1-q)\|=\max\{\|q\|,\|(a-1)(1-p)\|\}=\|q\|=1, $$ since $\|(a-1)(1-p)\|\leq\|a-1\|\,\|1-p\|\leq1$.
$\impliedby)$ This only works when $a$ is injective (see at the botoom for a counterexample when $a$ is not injective). Let $q\leq p$. Then $\|a-(1-q)\|=1$. Assume that $a$ is injective. Since $a-(1-q)$ is selfadjoint, its norm is achieved on the numerical range. That is, there exists $x$ with $\|x \|=1$ such that $$\tag1 \langle (a+q-1)x,x\rangle=\pm1. $$ When equal to $-1$, $(1)$ becomes $$ \langle ax,x \rangle+\langle qx,x \rangle=0. $$ This would imply that $ax=0$, contradicting that $a$ is injective. So we have $$ \langle ax,x \rangle+\langle qx,x \rangle=2. $$ As both summands are at most $1$, it follows that $$ \langle ax,x \rangle=1,\qquad\qquad \langle qx,x \rangle=1. $$ The last equality reads $\|qx\|=1$, and we deduce that $qx=x $. This allows us to write $\langle qaqx,x\rangle=1$, and we conclude that $$\tag2\|qaq\|=1,\qquad\qquad q\leq p.$$ Now fix $\alpha\in(0,1)$. Take $q=1_{[0,\alpha]}(pap)$. Then $q\leq p$, and $$ qaq=aq \leq\alpha q. $$ Then $qaq=0$, as otherwise we would have $\|qaq \|\leq\alpha<1$, a contradiction. This shows that $pap=1_{\{1\}}(pap)$, implying that $pap$ is a projection. We have $$ pa^2p\leq pap=(pap)^2=papap\leq pa^2p. $$ Thus $pa^2p=papap$, which we can write as $pa(1-p)ap=0$. This in turn implies $(1-p)ap=0$, so $ap=pap$ and $ap=pa$. In particular $ap=pap\leq p$. If $q=p-ap\ne 0$, we get $q\leq p$ and $$ qaq=q (ap)q=(p-ap)ap(p-ap)=0, $$ contradicting $(2)$. So $ap=p$, and hence $p\leq a$.
When $a$ is not injective, the reverse implication can fail. For example let $M=M_3(\mathbb C)$, and $$ a=\begin{bmatrix} 1&0&0\\0&0&0\\0&0&0\end{bmatrix}, \qquad\qquad p=\begin{bmatrix} 1&0&0\\0&1&0\\0&0&0\end{bmatrix} $$ Any $q\leq p$ is of the form $$ q=\begin{bmatrix} t&\sqrt{t-t^2}&0\\\sqrt{t-t^2}&1-t&0\\0&0&0\end{bmatrix}. $$ Then $$ a-(1-q)=\begin{bmatrix} t&\sqrt{t-t^2}&0\\\sqrt{t-t^2}&-t&0\\0&0&-1\end{bmatrix} $$ and hence $$ \|a\|=1,\qquad\qquad p\not\leq a,\qquad \qquad \|a-(1-q)\|=1\ \ \ \forall q\leq p. $$ This situation can be achieved for most projections $p$ in most von Neumann algebras. Namely, given a non-trivial, non-minimal projection $p$, let $q_0\leq p$ be a proper subprojection, and put $a=q_0$. Then $p\not\leq a$ and for any $q_1\leq p$ we have $$ a-(1-q_1)=q_0+q_1-1. $$ For any $x$ with $(1-p)x=x$, we have $(a-(1-q_1)x=-x$, and so $\|a-(1-q_1)\|=1$.