I think it's also instructive to look at this via the multiplication operator version of the spectral theorem. I'm not sure offhand of a canonical reference for this statement (can anyone supply one?) so I'll give it here.
Definition. Let $(X,\mu)$ be a measure space, and $\phi : X \to \mathbb{R}$ a measurable function. The multiplication operator $M_\phi$ corresponding to $\phi$ is the (possibly) unbounded operator on $L^2(X)$ with domain $D(M_\phi) = \{ f \in L^2(X) : \int_X |\phi(x) f(x)|^2\,\mu(dx) < \infty\}$ and defined by $(M_\phi f)(x) = \phi(x) f(x)$.
It is a simple exercise to show that:
It is also easy to see how the functional calculus works for such operators: if $g : \mathbb{R} \to \mathbb{R}$ is measurable, then $g(M_\phi) = M_{g \circ \phi}$.
Theorem (Spectral Theorem). Suppose $T$ is a self-adjoint operator on a Hilbert space $H$. There exists a measure space $(X,\mu)$, a measurable function $\phi : X \to \mathbb{R}$, and a unitary operator $U : H \to L^2(X)$ such that:
$h \in D(T)$ iff $Uh \in D(M_\phi)$; and
For all $h \in D(T)$, $Th = U^{-1} M_\phi U h$ (i.e. $T = U^{-1} M_\phi U$).
Informally, this says that, up to unitary equivalence, any self-adjoint operator is a multiplication operator. This also gives a functional calculus: $g(T) = U^{-1} M_{g \circ \phi} U$.
Now back to your problem. Thanks to this version of the spectral theorem, it suffices to handle the case when $T$ is a multiplication operator $M_\phi$ on some measure space $(X,\mu)$. Since $T$ is nonnegative, $\phi \ge 0$ a.e. Now we are trying to estimate $\|1_{[d,\infty)}(T) u\|^2 = \int_X |1_{[d,\infty)}(\phi(x)) u(x)|^2\,\mu(dx) = \int_X 1_{\{\phi \ge d\}}\,u^2\,d\mu = \nu(\{\phi \ge d\})$ where $\nu$ is the measure $\nu(B) = \int_B u^2\,d\mu$.
On the other hand, $\langle Tu , u \rangle = \int_X \phi u^2\,d\mu = \int_X \phi \,d\nu.$
So we can rewrite the statement we want to prove as: $\nu(\{\phi \ge d\}) \le \frac{1}{d} \int_X \phi\,d\nu.$ But this is nothing but Markov's inequality!