The square root is operator monotone

Step 1: positive definite case. Assume $0<A\le B$ and let $X=\sqrt B-\sqrt A$, a symmetric matrix. We have the algebraic identity
$$\sqrt B\,X+X\sqrt A=B-\sqrt B\sqrt A+\sqrt B\sqrt A-A=B-A.$$
Suppose $X$ had a negative eigenvalue $\lambda<0$ with eigenvector $v\ne0$. Since $X$ is symmetric, $v^TX=\lambda v^T$ and $Xv=\lambda v$. Then
$$v^T(B-A)v=v^T\sqrt B\,Xv+v^TX\sqrt A\,v=\lambda\,v^T\sqrt B\,v+\lambda\,v^T\sqrt A\,v=\lambda\big(v^T\sqrt Bv+v^T\sqrt Av\big)<0,$$
because $\sqrt A$ and $\sqrt B$ are positive definite so the bracket is positive, while $\lambda<0$. This contradicts $B-A\ge0$. So all eigenvalues of $X$ are $\ge0$, i.e. $\sqrt A\le\sqrt B$.

Step 2: general case. If $0\le A\le B$, then for each $\varepsilon>0$, $0<A+\varepsilon I\le B+\varepsilon I$, so by Step 1
$$\sqrt{A+\varepsilon I}\le\sqrt{B+\varepsilon I}.$$
The square root is continuous on the cone of positive semidefinite matrices, and the set $\{(X,Y):X\le Y\}$ is closed. Letting $\varepsilon\to0$ gives $\sqrt A\le\sqrt B$.

Example. The analogous statement for squares is false: $A=\begin{pmatrix}1&1\\1&1\end{pmatrix}\le B=\begin{pmatrix}2&1\\1&1\end{pmatrix}$ but $A^2=\begin{pmatrix}2&2\\2&2\end{pmatrix}\not\le B^2=\begin{pmatrix}5&3\\3&2\end{pmatrix}$, since $B^2-A^2=\begin{pmatrix}3&1\\1&0\end{pmatrix}$ has determinant $-1<0$.

Remark. More generally $t\mapsto t^r$ is operator monotone for $0\le r\le1$ (L\"owner-Heinz inequality) and not for $r>1$. This result is one of the first examples where order on matrices behaves differently from order on numbers.