Square root is operator monotone

If $X$ and $Y$ are $n\times n$ positive semidefinite matrices over $\mathbb{C}$ or $\mathbb{R}$, we can prove your statement as follows.

We first assume that $X$ is positive definite. Let $A=YX^{-1}$. Since $A$ is similar to $X^{-1/2}YX^{-1/2}$, all eigenvalues of $A$ are positive. Let $v$ be a unit eigenvector corresponding to $\lambda_\min(A)$. As $X^2\le Y^2$, we have $I \le X^{-1}Y YX^{-1} = A^TA$ and hence $1\le v^TA^T Av=\lambda_\min(A)^2$. Thus all eigenvalues of $X^{-1/2}YX^{-1/2}\sim A$ are bounded below by $1$, i.e. $I\le X^{-1/2}YX^{-1/2}$. Hence $X\le Y$.

Now the positive semidefinite case can be obtained as a limiting case of positive definite cases.

Afternote: In the above proof, the only irreversible step is $I\le A^TA\Rightarrow1\le\lambda_\min(A)$. This is because we have $\sigma_\min(A)\le|\lambda|_\min(A)$ in general, but strict inequality can hold. The irreversibility of this step indicates that, while the square root function is operator monotone in our case, the square function is not. (Thanks for julien for pointing out that.)