Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $(F,\langle\cdot,\cdot\rangle)$. Let $M\in \mathcal{B}(F)^+$ (i.e. $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$).
Why for any $x\in F$ there exists a sequence $(x_n)_n$ with $$M^{1/2}x=\lim_{n\to \infty} Mx_n\;?$$
On
Let $M=\int \lambda \,\mathrm{d}E_\lambda$ be the spectral decomposition. Let $x_n=\int_{\lambda>1/n}\lambda^{-1/2}E_{\lambda}(x)\,\mathrm{d}E_\lambda$. Then for any $y$ you have $$\langle Mx_n\mid y \rangle=\int_{\lambda>1/n}\langle \lambda^{1/2}E_{\lambda}(x)\mid y \rangle\,\mathrm{d}E_\lambda=\langle M^{1/2} x\mid y\rangle-\int_{\lambda\leq 1/n}\langle \lambda^{1/2}E_{\lambda}(x)\mid y\rangle \,\mathrm{d}E_\lambda.$$
But $\lvert\langle \lambda^{1/2}E_\lambda(x)\mid y\rangle\rvert\leq\lambda^{1/2}\lVert x\rVert\lVert y\rVert$, so for fixed $x$ and $\lVert y\rVert\leq 1$ it tends to zero uniformly (in $y$) as $\lambda\to 0$. The conclusion follows.
More directly, $Mx_n-M^{1/2}x=\int_{\lambda\leq 1/n}\lambda^{1/2}E_\lambda(x) \,\mathrm{d}E_\lambda$, and likewise $\lVert \lambda^{1/2}E_\lambda(x)\rVert\leq \lambda^{1/2}\lVert x\rVert$.
Define a sequence of polynomials $$p_0(t) = 0$$ $$p_{n+1}(t) = p_n(t) + \frac12\left(t - p_n(t)^2\right), \quad\forall n \in \mathbb{N}$$
This answer shows that $p_n(M) \xrightarrow{n\to\infty} M^{1/2}$ in the strong operator topology.
Notice that for all $n \in \mathbb{N}$ we have $p_n(0) = 0$ so we can write $p_n(t) = tq_n(t)$. Now for any vector $x$ we have
$$M^{1/2}x = \lim_{n\to\infty} p_n(M)x = \lim_{n\to\infty} M(q_n(M)x)$$
which is the desired claim.