Let $\{E(\lambda)\}_{\lambda \in \mathbb{R}}$ be a spectral family on a Hilbert space $\mathcal{H}$, meaning a collection of orthogonal projectors satisfying, for any $f \in \mathcal{H}$
$ \quad (i) \ \lim \limits_{\lambda \to -\infty} E(\lambda)f = 0 \quad $ and $\quad \lim \limits_{\lambda \to +\infty} E(\lambda)f = f$
$ \quad (ii) \ \ (E(\lambda)f,f) \le (E(\mu)f,f) \ \ $ for $\lambda \le \mu$
$ \quad (iii) \ \lim \limits_{\epsilon \to 0^+} E(\lambda + \epsilon)f = E(\lambda)f$
Then, $E(\lambda)E(\mu) = E(\mu)E(\lambda) = E(\lambda) \ \ $ for $\lambda \le \mu$
My attempt
Since $E(\lambda)$ is a projector, $f \in \mathrm{Ran} \ E(\lambda) \iff f = E(\lambda)f$. $\\$ Also, from $(ii)$, one gets $\mathrm{Ran} \ E(\lambda) \subseteq \mathrm{Ran} \ E(\mu)$. Using these two: since obviously $E(\lambda)f \in \mathrm{Ran} \ E(\lambda) \subseteq \mathrm{Ran} \ E(\mu)$, it follows $E(\mu)(E(\lambda)f) = E(\lambda)f$, so $E(\mu)E(\lambda) = E(\lambda)$.
I am however having trouble proving the other half of the statement, $E(\lambda)E(\mu) = E(\lambda)$. Maybe I am missing something obvious
Upon request, here is the answer contained in my comment above:
Once the OP has already proven that $E(\mu)E(\lambda) = E(\lambda)$, and observing that orthogonal projectors are self adjoint, we have $$ E(\lambda)E(\mu) = \big(E(\mu)E(\lambda)\big) ^* = E(\lambda)^* = E(\lambda).$$