If $B$ is a $C^*$-subalgebra of $A$, then $sigma_A(b) cup {0} = sigma_B(b) cup {0}$

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
$$sigma_B(b) cup {0} stackrel{(1)}= sigma_{B oplusBbb{C}1_A}(b) cup {0}$$
$$stackrel{(2)}=sigma_A(b) cup {0}$$

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?