Prove that for self-adjoint operators $T,P \in B(H)$ where $H$ is a Hilbert space ($B(H)$ is the space of all bounded functions from $H \to H$) and $T \leq P$ then $\|T\|\leq \|P\|$ .
So by assumption $T\leq P$ so $\langle Tx,x \rangle \leq \langle Px,x \rangle$ for every $x \in H$. So then the norm is just the supremum over all $x\in H$ where $\|x\| = 1$ and this holds because the inequality was true for every $x\in H$. Is this correct? I feel like it is quite obvious.
We need additional assumption to make your statement as true. A very obvious counterexample is $T=-2I$, $P=I$.
If we make additional assumption that $O\le T\le P$, i.e. $T,P$ are non-negative elements of $B(H)$, then it is true that $\|T\|\le \|P\|$. Here's a spectral argument that can show this. Note that the norm is equal to the spectral radius for every self-adjoint operator. Let $\lambda = \|P\|$ be the maximal element of the spectrum $\sigma(P)$. Since $\sigma(P)\subset [0,\lambda]$, we have $O\le T\le P\le \lambda I$. We find that $ \sigma(T)\subset [0,\lambda], $ which implies that $\|T\|\le \lambda = \|P\|$.