= Solution
A <Hermitian matrix> satisfies $A=A^\dagger$, where $\dagger$ means conjugate transpose. If $Av=\lambda v$ with $v\neq0$, then $v^\dagger Av=\lambda\|v\|^2$ is real, since it equals its complex conjugate $v^\dagger A^\dagger v$. Thus every <eigenvalue> is real. To prove the finite-dimensional <spectral theorem>, choose a unit <eigenvector> $v$; one exists because the complex <characteristic polynomial> has a root. Its <orthogonal complement> is invariant, since for $w\perp v$,
$$
v^\dagger Aw=(Av)^\dagger w=\lambda v^\dagger w=0.
$$
The restriction to that complement is again Hermitian. Induction on dimension supplies an <orthonormal basis> of <eigenvectors>, so $A=U\operatorname{diag}(\lambda_1,\ldots,\lambda_n)U^\dagger$ for a <unitary matrix> $U$.
In this basis, $x^\dagger Ax=\sum_i\lambda_i|(U^\dagger x)_i|^2$. Therefore $\boxed{A>0\iff\lambda_i>0\text{ for every }i}$. In that case the <square root of a matrix>
$$
C=U\operatorname{diag}(\sqrt{\lambda_1},\ldots,\sqrt{\lambda_n})U^\dagger
$$
is Hermitian, positive definite and satisfies $C^2=A$. For uniqueness, any positive-definite Hermitian $D$ satisfying $D^2=A$ commutes with $A$ and hence preserves each <eigenspace> of $A$. Its restriction to the $\lambda$-eigenspace is Hermitian, and its positive <eigenvalues> all solve $\mu^2=\lambda$. Thus that restriction is $\sqrt\lambda I$, forcing $D=C$. This is the <principal square root of a positive semidefinite matrix>, here strictly positive definite.
Now let $D=\sqrt B$ and $X=C-D$. Expanding without assuming that $C$ and $D$ commute gives
$$
CX+XC=2C^2-CD-DC=(A-B)+X^2.
$$
The <Hermitian matrix> $X$ has $X^2\geq0$ because $v^\dagger X^2v=\|Xv\|^2$, whereas $A-B>0$. Hence $CX+XC>0$. If $Xv=\mu v$, $v\neq0$, then
$$
0<v^\dagger(CX+XC)v=2\mu\,v^\dagger Cv.
$$
Since $C>0$, every real <eigenvalue> $\mu$ of $X$ is positive. The positive-eigenvalue criterion now gives the strict finite-dimensional <operator monotonicity of the square root>:
$$
\boxed{CX+XC>0,\qquad \sqrt A-\sqrt B=X>0.}
$$
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