Let $a,b$ self-adjoint elements in a unital $C^*$-algebra. I'm trying to show that $$0 \leq a \le b \implies \sqrt{a} \le \sqrt{b}$$ (these square roots are defined via functional calculus).
The hint I was given said to prove it first if $b$ is invertible, and I managed to do that.
Next, I attempted the case that $b$ is non-invertible. Fix $\epsilon > 0$. A standard argument using Gelfand's theorem shows that $b + \epsilon$ is invertible, and we have $a \le b + \epsilon$,so the former case applies and we get
$$\forall \epsilon > 0: \sqrt{a} \leq \sqrt{b+ \epsilon}$$
Next, choose a sequence $\epsilon_n \searrow 0$. I want to show that
$$\sqrt{b+ \epsilon_n} \to \sqrt{b} \quad\quad (*)$$ Maybe this follows because $f_n \to f$ uniformly where $f_n(x) = \sqrt{x+\epsilon_n}$ and $f(x) = \sqrt{x}$ (via functional calculus)?
If I can deduce this, then from $$\forall n : \sqrt{a} \leq \sqrt{b+\epsilon_n}$$ we would obtain $\sqrt{a} \leq \sqrt{b}$ where implicitely was used that the positive elements are norm-closed.
So how can I justify $(*)$. Is the reason I said a good one?

The reason is that functional calculus is a $*$-homomorphism. Concretely, let $\Gamma:C(\sigma(b))\to C^*(b)$ be the Gelfand transform. Then $$ \|f(b)-g(b)\|=\|\Gamma(f)-\Gamma(g)\|=\|\Gamma(f-g)\|\leq\|f-g\|. $$