Hypothesis: Assume, that $A$ and $B$ are positive bounded operators (on some Hilbert space $H$) and $A\geq B \geq 0$. Then ${\rm range}(A) \supset {\rm range}(B)$.
The textbook "$C^*$-algebras by example" seems to be using this hypothesis.
Is the hypothesis true? How to prove it?
There is an easier way to prove the (weaker) result of the other answer (by Vahid Shirbisheh). Recall that every operator $T \in B(H)$ satisfies $$ \overline{\text{range}(T)} = \ker(T^*)^\perp. $$ In particular, for self-adjoint operators such as $A$ and $B$ we have \begin{align*} \overline{\text{range}(A)} = \ker(A)^\perp,\\[1ex] \overline{\text{range}(B)} = \ker(B)^\perp. \end{align*} We assume that $0 \leq B \leq A$ holds. Equivalently: for all $x\in H$ we have $0 \leq \langle Bx,x\rangle \leq \langle Ax,x\rangle$. Thus, for $x\in \ker(A)$ we have $$ \lVert B^{1/2}x\rVert^2 = \langle B^{1/2}x,B^{1/2}x\rangle = \langle Bx,x\rangle \leq \langle Ax,x\rangle = 0, $$ hence $Bx = B^{1/2}B^{1/2}x = B^{1/2}0 = 0$. It follows that $\ker(A) \subseteq \ker(B)$ holds. Taking orthogonal complements, we find $$ \overline{\text{range}(A)} \supseteq \overline{\text{range}(B)}.\tag*{$\blacksquare$} $$
Note that the original question remains unanswered: it is still unclear whether $\text{range}(A) \supseteq \text{range}(B)$ holds, without taking closures. As a matter of fact, the answer to this question is no.
Counterexample. Let $H$ be the sequence space $\ell^2(\mathbb{N}^+)$ and let $T \in B(H)$ be an operator that is positive, injective and compact (for instance, $T$ could be the diagonal operator $e_n \mapsto \frac{1}{n}e_n$). Then $T$ is not surjective, so we may choose some unit vector $x_0\in H \setminus\text{range}(T)$.
Let $P$ be the orthogonal projection onto $\text{span}(x_0)$. Now we have $0 \leq P \leq P + T$ and $x_0\in\text{range}(P)$. Suppose, for the sake of contradiction, that $x_0\in\text{range}(P + T)$ holds. Then we may choose some $y\in H\setminus\{0\}$ such that $(P + T)y = x_0$. But now we have $Ty = x_0 - Py \in \text{span}(x_0)$. Since $T$ is injective, we also have $Ty \neq 0$, but this contradicts our assumption that $x_0 \notin \text{range}(T)$. We conclude that $\text{range}(P) \not\subseteq \text{range}(P + T)$; we only have $$ \text{range}(P) \subseteq \overline{\text{range}(P)} \subseteq \overline{\text{range}(P + T)}.\tag*{$\boxtimes$} $$
If $A$ and $B$ are projections, then it is easy to prove the statement. So, we reduce the problem to this case. Of course, I can only show that the closure of the image of $B$ is a subset of the closure of the image of $A$.
Let $P_A$ and $P_B$ be the support projections associated with $A$ and $B$, that is projections on the closure of the range of $A$ and $B$, respectively. Since $A$ and $B$ are positive we have $A^\alpha\to P_A$ and $B^\alpha \to P_B$ strongly as $\alpha\to 0$, see page 21 of Blackadar's book. Combining the above limits and the inequality $B^\alpha\leq A^\alpha$ for all $0<\alpha<1$ and the fact that strong convergence preserves the inequality between positive elements, we get $P_B\leq P_A$. Therefore $\overline{Im(B)}=Im(P_B)\subseteq Im(P_A)= \overline{Im(A)}$.
