= Solution
A bounded <self-adjoint operator> $T$ is a <positive operator> when $\langle T\xi,\xi\rangle\geq0$ for every $\xi$. Equivalently its <quadratic form> is nonnegative. We construct its <positive square root of an operator> directly and prove uniqueness.
If $T=0$ its positive square root is zero. Otherwise set $r=\|T\|$, $B=T/r$, and $C=I-B$. Then $0\leq B\leq I$ and $C$ is positive. It is also a contraction. Indeed, the <Cauchy-Schwarz inequality> for the positive form $\langle B\xi,\eta\rangle$ gives $\|B\xi\|^2\leq\langle B\xi,\xi\rangle$, and hence
$$
\|C\xi\|^2=\|\xi\|^2-2\langle B\xi,\xi\rangle+\|B\xi\|^2\leq\|\xi\|^2.
$$
The scalar binomial series has the form
$$
\sqrt{1-z}=1-\sum_{n\geq1}c_nz^n,\qquad
c_n=\frac{\binom{2n}{n}}{4^n(2n-1)}>0,\qquad\sum_{n\geq1}c_n=1.
$$
The last equality follows by taking $z\uparrow1$ in the nonnegative series. Therefore
$$
S=\sqrt r\left(I-\sum_{n\geq1}c_nC^n\right)
$$
converges in <operator norm>. Every $C^n$ is a positive contraction: its even powers are squares, its odd powers have form $C^kCC^k$, and their <norms> are at most one. Each partial sum for $S/\sqrt r$ is consequently positive, since it is at least $(1-\sum_{n\leq N}c_n)I$. The limit is positive. Multiplying the absolutely convergent series and using the scalar coefficient identity gives $S^2=r(I-C)=T$. This construction also shows that $S$ is a <norm> limit of polynomials in $T$ and thus commutes with every operator commuting with $T$.
Now suppose $R$ is another positive square root. It commutes with $T=R^2$, hence with $S$. Thus
$$
(R-S)(R+S)=R^2-S^2=0.
$$
The difference vanishes on the range of $R+S$ and on its <closure>. It also vanishes on its <kernel>: if $(R+S)\xi=0$, the fact that $R$ and $S$ are <positive operators> gives $\langle R\xi,\xi\rangle=\langle S\xi,\xi\rangle=0$, and the <Cauchy-Schwarz inequality> for their positive forms implies $R\xi=S\xi=0$. Since $R+S$ is <self-adjoint>, the orthogonal sum of its <kernel> and closed range is all of $H$. Therefore $R=S$, proving
$$
\boxed{T\geq0\Longrightarrow\text{a unique }T^{1/2}\geq0\text{ with }(T^{1/2})^2=T.}
$$
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