Suppose we are given a C*-algebra $\mathcal{X}$ that has no unit. Consider the unitization $$\mathcal{X}_1 \equiv \{x + \lambda \mid x \in \mathcal{X}, \lambda \in \mathbb{C}\}$$ of $\mathcal{X}$, endowed with the natural operations. Define a norm on $\mathcal{X}_1$ by $\|x+\lambda\| \equiv \sup \{ \|xa + \lambda a\| \colon a \in \mathcal{X}, \|a\| \leq 1\}$. How can one prove that $\mathcal{X}_1$ is complete with respect to this new norm?
Take a Cauchy sequence in $\mathcal{X}_1$, say $\{x_n + \lambda_n\}_n$. Then $\forall \epsilon>0 \exists N(\epsilon)$ such that for all $n,m \geq N(\epsilon)$ and for all $a \in \mathcal{X}, \|a\| \leq 1$ we have $\|(x_n-x_m)a + (\lambda_n-\lambda_m)a\| < \epsilon$. From that we know $\{x_n a + \lambda_n a\}_n$ is Cauchy in $\mathcal{X}$ for all $a \in \mathcal{X}, \|a\| \leq 1$. Hence for each $a \in \mathcal{X}, \|a\| \leq 1$ there exists $y(a) \in \mathcal{X}$ with $x_n a + \lambda_n a \to y(a)$ by completeness of $\mathcal{X}$. At this point I am stuck, since I don't quite see how to get an $\epsilon$-bound for $\|x_n + \lambda_n - y\|$.
Can anyone give hints on how to proceed?
Note that $\mathcal{X}_1 =\mathcal{X}\oplus \mathbb{C}$ as $\mathbb{C}$-vector spaces.
We also have an isometric inclusion:
$$i: \mathcal{X} \to \mathcal{X}_1: a \mapsto a+0$$
Indeed, for $a \neq 0$
$$\Vert L_a\Vert = \sup_{\Vert x \Vert \leq 1} \Vert a x \Vert \leq \Vert a\Vert = \frac{\Vert aa^*\Vert}{\Vert a \Vert} = \frac{\Vert L_a a^* \Vert}{\Vert a \Vert} \leq \Vert L_a \Vert \frac{\Vert a^*\Vert}{\Vert a \Vert }= \Vert L_a \Vert$$
and thus $\Vert i(a) \Vert = \Vert L_a \Vert = \Vert a \Vert$. Note that the $C^*$-structure was used to justify the second equality above.
Thus, we see that $\mathcal{X_1}/i(\mathcal{X}) \cong \mathbb{C}$ is a finite-dimensional vector space.
Now, use the following result to conclude: