Square of an approximate unit is an approximate unit Let $A$ be an abelian $C^*$-algebra with approximate unit $(u_\lambda)$. I'm trying to show that $(u_\lambda^2)$ is an approximate unit as well. In particular, I'm stuck at showing that
$$\lim_\lambda \| a-u_\lambda^2 a \| = 0$$
for all $a \in A$.
Here is my approach:
Since $A$ is spanned by its positive elements, we may suppose that $a\geq 0$. We have
$$\lim_\lambda u_\lambda^2 a = \lim_\lambda u_\lambda^2 \sqrt{a}^2 = \left(\lim_\lambda u_\lambda \sqrt{a}\right)^2 = \sqrt{a}^2 = a$$
and the result then follows. Note that commutativity was used for the rule $(ab)^2 = a^2b^2$ to hold.
Is the above correct? Can we drop the assumption that $A$ is abelian?

Edit: initially thought it was true for general C*-algebras. As pointed out in the comments, I forget about $\lambda_1 \leq \lambda_2 \Rightarrow u_{\lambda_1} \leq u_{\lambda_2}$. To remedy this, assuming abelian fixes it since $0 \leq x \leq y \Rightarrow x^2 \leq y^2$ in the abelian case, but not in general.
Let $A$ be any C*-algebra, abelian or not. Then if $(u_\lambda)$ is an approximate unit, these are all positive contractions, and so their squares are as well. To see that the limits are as desired, notice that
$$ \|a - u_\lambda^2a\| \leq \|a - u_\lambda a\| + \|u_\lambda a - u_\lambda^2 a\| \leq \|a - u_\lambda a\| + \|u_\lambda\|\|a - u_\lambda a\|$$
where the first inequality just follows by adding and subtracting $u_\lambda a$ and using the triangle inequality and the second comes from the submultiplicativity of the norm. From here its clear that the right hand side tends to 0 since $(u_\lambda)$ is an approximate unit (the norm differences tend to zero and $\|u_\lambda\| \leq 1$).
Now assuming that $A$ is abelian, $\lambda_1 \leq \lambda_2 \Rightarrow u_{\lambda_1} \leq u_{\lambda_2} \Rightarrow u_{\lambda_1}^2 \leq u_{\lambda_2}^2$, so that $(u_{\lambda}^2)$ is an approximate identity.
