$\Vert x \Vert \leq 1 \implies \Vert x^{1/2} \Vert \leq 1$ in a $C^*$-algebra

Question

Let $A$ be a $C^*$-algebra and $x \in A$ a positive element with $\Vert x \Vert \leq 1$. Is it true that $\Vert x^{1/2}\Vert \leq 1?$

Attempt: Note that $C^*(x)$ is an abelian $C^*$-algebra, so we can consider the Gelfand representation $$\phi: C^*(x) \to C_0(\Omega)$$ where $\Omega$ is a locally compact Hausdorff space. Note that $\phi$ preserves (positive) square roots of positive elements because these are unique. Put $f:= \phi(x)$. Then $\Vert f^{1/2}\Vert_\infty \leq 1$ holds trivially so also $\Vert x^{1/2}\Vert \leq 1$. Hence, we are done.

Is the above correct?

Answer 1

$x^{1/2}$ is also positive and in particular self-adjoint so by $C^*$-property we have $$\left\|x^{1/2}\right\|^2 = \left\|\left(x^{1/2}\right)^*x^{1/2}\right\| = \|x\| \le 1.$$

Answer 2

Yes this is correct. Perhaps you need also to explain that every element $x^{1/2}$ (there may be several a-priory) belongs to $C^*(x)$.

Edit. Sorry I forgot that the positive square root of a positive element in a $C^*$-algebra is unique and belongs to $C^*(x)$. But it would be better if you refer to that result. It is not completely obvious.