proof that this is an isometric map (on a $C^*$-module)

Question

Are my steps right? I'm not sure about the statement in bold below.

Let $A$ be a $C^*$-algebra. Let $X$ be an $A$-module.

Let $x\in X$, let $a= \langle x,x \rangle $ Define $\lambda _a (z) = az$, $z\in X$.

I want to show that $\| \lambda _a \| = \| a \| $

Let $e_\alpha = \langle u_\alpha , v_\alpha \rangle $ be an approximate identity for the $C^*$-algebra $\langle X,X\rangle $, such that $u_\alpha $ and $v_\alpha $ are norm bounded by $1$. ( is this statement true? I know that all $C^*$-algebras have bounded approximate identities)

it is easy to see that $\| \lambda _a \| \leq \| a \| $

Moreover,

\begin{align*} \| ae_\alpha \| &= \| \langle au_\alpha , v_\alpha \rangle \| = \| \langle \lambda _a (u_\alpha ),v_\alpha \rangle \| \\ & \leq \| \lambda _a (u_\alpha ) \| \| v_\alpha \| \leq \| \lambda _a \| \end{align*}

Therefore, $\| a \| \leq \| \lambda _a \| $ and hence, $\| a \| = \| \lambda _a \| $

Answer 1

To see that $\|\lambda_a\|\geq \|a\|$, consider $\|\lambda_a(x)\|$.