Bound operator norm Consider the Hilbert space $\ell^2(\mathbb{N})$ and let $P:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N})$ be an orthogonal projection. Let $(\delta_i)_{i\in\mathbb{N}}$ denote the canonical orthonormal basis of $\ell^2(\mathbb{N})$ and define for $C\subset\mathbb{N}$
$$
Q_C=\sum_{k\in C}Q_k,
$$
where $Q_kx=\langle\delta_k,x\rangle\delta_k$. Put differently, $Q_C$ is the orthogonal projection on $\ell^2(C)=\overline{\mathrm{span}\{\delta_k\}_{k\in C}}$. 
Now I want to prove:
$$
\|Q_CP(\mathrm{Id}-Q_C)PQ_C\|\leq \frac14.
$$
My attemp: By the min-max theorem, $\|Q_CP(\mathrm{Id}-Q_C)PQ_C\|=\sup_x\langle x,(Q_CP(\mathrm{Id}-Q_C)PQ_C)x\rangle$ for $\|x\|=1$. By the outermost $Q_C$'s, we may restrict from the beginning to $x\in\overline{\mathrm{span}\{\delta_k\}_{k\in C}}$. In summary,
$$
\|Q_CP(\mathrm{Id}-Q_C)PQ_C\|=\sup_{\substack{x\in\ell^2(C)\\\|x\|=1}}\langle x,P(\mathrm{Id}-Q_C)Px\rangle=\sup_{\substack{x\in\ell^2(C)\\\|x\|=1}}\left[\|Px\|^2-\sum_{k\in C}|\langle Px,\delta_k\rangle|^2\right].
$$
However, this approach does not look very fruitful...
 A: I give the argument based on Halmos' paper as suggested by Julien in his comment. All credits are due to him for pointing out this neat result.
Let us first cite the theorem of Halmos:

If $M$ and $N$ are subspaces in generic position in a Hilbert space $H$, with respective projections $\tilde{P}$ and $Q$, then there exists a Hilbert space $K$, and there exist positive contractions $S$ and $C$ on $K$, with $S^2 + C^2 = 1$ and $\ker S=\ker C=0$, such that $\tilde{P}$ and $Q$ are unitarily equivalent to
  $$
\begin{pmatrix}1& 0\\0&0\end{pmatrix},\qquad \begin{pmatrix}C^2&CS\\CS&S^2\end{pmatrix}
$$
  respectively.

What's not written in the theorem but follows by the proof is that $K=\mathrm{ran}P$ and that $\tilde{P}$ and $Q$ admit the same unitary $U:H\to K\oplus K$. Moreover, $C$ and $S$ commute. Then using Halmos' theorem with $\tilde{P}=Q_C$, $Q=P$, we see by a little computation
$$
\|Q_CP(\mathrm{Id}-Q_C)PQ_C\|=\left\|\begin{pmatrix}C^2S^2&0\\0&0\end{pmatrix}\right\|=\|C^2-C^4\|.
$$
An easy exercise in spectral calculus shows $\sigma(C^2-C^4)\subset [0,1/4]$ from which the result follows. 
A: The way I would see your inequality is as 
$$
\|(I-Q_C)PQ_C\|\leq\frac12,
$$
since $$\|Q_CP(I-Q_C)PQ_C\|=\|Q_CP(I-Q_C)^*(I-Q_C)PQ_C\|=\|(I-Q_C)PQ_C\|^2.$$
And this is basically saying that the off diagonal entries of a projection cannot contribute more than $1/2$ to the norm. In spirit, this happens because the equality $P^2=P$ gives, for every diagonal entry, $$ P_{kk}-P_{kk}^2=\sum_{j\ne k} |P_{kj}|^2.$$
To make this into a proof, we write $P$ in terms of the decomposition $H=Q_CH\oplus (I-Q_C)H$, 
$$
P=\begin{bmatrix} A&B\\ B^*&C\end{bmatrix},
$$
with $A,C\geq0$ (since $P$ is positive). If we now look at the equality $P^2=P$ at the $1,1$ entry, we get 
$$
A^2+BB^*=A.
$$
So $BB^*=A-A^2$. As $A$ is a positive contraction (from $P$ being a projection), we have 
$$
\sigma(A-A^2)\subset\{\lambda-\lambda^2:\ \lambda\in[0,1]\}.
$$
For all such $\lambda$ we have $\lambda-\lambda^2\leq1/4$, so $$\|B\|^2=\|BB^*\|=\|A-A^2\|\leq\frac14,$$ so $\|B\|\leq1/2$. Thus
$$
\|(I-Q_C)PQ_C\|=\|B\|\leq\frac12.
$$
