Suppose $0\leq A\leq B$ (and hence $A$ and $B$ are self-adjoint) where $A$ and $B$ are compact on a complex Hilbert space $H$. That is, $$0\leq \langle x,Ax\rangle\leq \langle x,Bx\rangle$$ for all $x\in H$.
Is it true that $0\leq A^2\leq B^2$?
Since $A$ and $B$ are positive and self-adjoint, by the spectral theorem there exist orthonormal bases of eigenvectors $(\varphi_n)$ and $(\psi_m)$ with corresponding non-negative eigenvalues $(\lambda_n)$ and $(\mu_m)$ for $A$ and $B$ respectively. That is, $A=\sum_n\lambda_n\langle{\varphi_n,\cdot}\rangle\varphi_n$ and $B=\sum_n\mu_m\langle{\psi_m,\cdot}\rangle\psi_m$.
So for any $x\in H$, $$\langle x,Ax\rangle = \langle x, \sum_n\lambda_n\langle{\varphi_n,x}\rangle\varphi_n\rangle=\sum_n\lambda_n|\langle \varphi_n,x\rangle|^2,$$ and similarly $$\langle x,Bx\rangle =\sum_m\mu_m|\langle \phi_m,x\rangle|^2.$$
The assumption $0\leq A\leq B$ implies that $$\sum_n\lambda_n|\langle \varphi_n,x\rangle|^2\leq \sum_m\mu_m|\langle \phi_m,x\rangle|^2$$.
But to compare $A^2$ and $B^2$ I must consider $$\langle x,A^2x\rangle =\sum_n\lambda_n^2|\langle \varphi_n,x\rangle|^2\quad\text{and}\quad \langle x,B^2x\rangle=\sum_m\mu_m^2|\langle \phi_m,x\rangle|^2$$and I'm not sure how to do so!
The answer is "no": Consider $$ A:=\left( \begin{array}{cc} 2 & 0 \\ 0 & 1 \end{array} \right), \quad B:=\left( \begin{array}{cc} 3 & -1 \\ -1 & 2 \end{array}\right). $$ Clearly $A \ge 0$, and since $$ B-A =\left( \begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right) $$ has the characteristic polynomial $$ (1-\lambda)^2 - 1 = \lambda^2 - 2 \lambda = \lambda(\lambda -2) $$ we have $A \le B$. Moreover $$ A^2 =\left( \begin{array}{cc} 4 & 0 \\ 0 & 1 \end{array}\right), ~ B^2=\left( \begin{array}{cc} 10 & -5 \\ -5 & 5 \end{array}\right); $$ $$ B^2 - A^2 =\left( \begin{array}{cc} 6 & -5 \\ -5 & 4 \end{array}\right), ~ {\rm det}\; (B^2 - A^2) = 24 - 25 = -1. $$ Thus $A^2 \not\le B^2$.