In Murphy's book on $C^*$-algebras, it is claimed that f $B$ is a $C^*$-subalgebra of a $C^*$-algebra $A$, then $\sigma_A(b) \cup \{0\} = \sigma_B(b) \cup \{0\}$.
Denote the unitalisation of $A$ by $\tilde{A}$.
To prove this, I will freely use the following results:
(1) If $A$ is a unital algebra and $B$ a subalgebra with $B + \Bbb{C}1_A = A$, then $\sigma_B(b) \cup \{0\} = \sigma_A(b) \cup \{0\}$ for $b \in B$.
(2) If $A$ is a unital $C^*$-algebra and $B$ is a $C^*$-subalgebra with $1_A \in B$, then $\sigma_A(b) =\sigma_B(b)$ for $b \in B$.
My attempt:
Case 1: $A$ is unital.
If $1_A \in B$, we are done by $(2)$. So suppose that $1_A \notin B$. Then we have $$\sigma_B(b) \cup \{0\} \stackrel{(1)}= \sigma_{B \oplus\Bbb{C}1_A}(b) \cup \{0\}$$ $$\stackrel{(2)}=\sigma_A(b) \cup \{0\}$$
Case 2: $A$ is non-unital. Then $$\sigma_A(b) \cup \{0\}=\sigma_{\tilde{A}}(b) \cup \{0\}$$
However, $\tilde{A}$ is a unital $C^*$-algebra and since $A$ is a $C^*$-subalgebra of $\tilde{A}$, we also have that $B$ is a $C^*$-subalgebra of $\tilde{A}$. Consequently, by case $(1)$ we get $$\sigma_{\tilde{A}}(b) \cup \{0\} = \sigma_B(b) \cup \{0\}$$
and we can conclude the proof. $\quad \square$
Is the above proof correct?