Let $A$ be a $C^*$-algebra and let $x\in A$. We can always embed $A\subset B(H)$ by taking the universal representation. By the polar decomposition of $B(H)$, we know that $x$ is written as $x=v|x|$ where $v$ is a partial isometry in $B(H)$. Moreover, we know that $|x|=v^*x$ and that if $x=w|x|$ for some other partial isometry that satisfies $\ker(w)=\ker(x)$ (when seen as operators), then $w=v$.
Actually, this is possible in any von Neumann algebra and not just $B(H)$, so the partial isometry $v$ explained above belongs to the strong closure of $A$ in $B(H)$, which is the enveloping von Neumann algebra of $A$, which by the Sherman-Takeda theorem is the double dual $A^{**}$. In other words, for a $C^*$-algebra $A$, we can do polar decomposition of its elements in $A^{**}$. So far so good.
My question is this: Let's say we are in the above setting, so $x\in A$, $x=v|x|$ where $v\in A^{**}$ is a partial isometry (and $|x|=v^*x$). I have seen remarked in a paper by Kirchberg that, if $y\in\overline{x^*Ax}$, then $vy\in A$. However, I cannot prove it. By elementary arguments it is enough to show that given $a\in A_{+,1}$ we have that $vx^*ax\in A$. Now if $a$ was of the form $xz$ for $z\in A$, then we would be done, since $vx^*ax=vx^*xzx=v|x|^2zx=x|x|zx\in A$. Is it something that obvious that I am missing here?
Also, as far as I know, the inclusion $A\subset A^{**}$ is only an inclusion of a $C^*$-algebra in a von Neumann algebra (irrelevant: with $\text{ball}(A)$ being weak-star dense in $\text{ball}(A^{**})$) ; I mean, $A$ is not necessarily a hereditary $C^*$-subalgebra of $A^{**}$ or anything like that, correct?
Here is an elementary answer I just came up with:
Set $B_x:=\{y\in A: vy\in A, vy^*\in A\}$. Note that it is very easy to verify that $B_x$ is a $C^*$-subalgebra of $A$. Moreover, $B_x$ is hereditary in $A$: indeed, if $b,b'\in B_x$ and $a\in A$, then $v(bab')=(vb)ab'\in A$, since $vb\in A$. Also, $v(bab')^*=(vb'^*)a^*b^*\in A$, since $vb'^*\in A$. Therefore $bab'\in B_x$, as we wanted.
Also, $x^*x\in B_x$: indeed, $v(x^*x)=v|x|^2=x|x|\in A$ and $x^*x$ is self-adjoint so yes, $x^*x\in B_x$.
But now if $a\in A_{+,1}$ then $0\leq a\leq 1$. Now $0\leq x^*ax\leq x^*x$, so $x^*ax\in B_x$ because $B_x$ is hereditary in $A$, which proves that $vx^*ax\in A$, as we wanted.