From Functional Analysis: Spectral Theory, by V. S. Sunder, page 97.
Let $A$ denote a unital commutative Banach algebra, then $\hat{x} (\hat{A}) = \sigma(x)$, where $\hat{A}$ is the collection of all complex homomorphisms from $A$ and for $x \in A$ and $\phi \in \hat{A}$, $\hat{x} : \hat{A} \to \mathbb{C}$ is defined by $\hat{x}(\phi)=\phi(x)$ and $\sigma (x)$ is the spectrum of $x$.
Proof. If $x ∈ A$, $\phi ∈ \hat{A}$, then $(x − \phi(x)1) ∈ \ker \phi$, and hence, by Proposition 3.2.4, we see that $\phi(x) ∈ σ(x)$. Conversely, it follows from Proposition 3.2.4 and (a) of this Lemma, that $σ(x) ⊂ \hat{x}(\hat{A})$.
I'm having trouble understanding why $(x − \phi(x)1) ∈ \ker \phi$. I get that this being true implies $\phi$ is not invertible, i.e. it is in $\sigma(x)$, but not $why$ it is true. I tried $$\begin{align*} \phi(x − \phi(x)1) &= \phi(x) - \phi(x)\phi(1)\\ &=\phi(x)-\phi(x)\\ &=0 \end{align*}$$ but I'm not sure my argument is correct. The second part of the proof is also obscure to me as it is very concise, so some help there would be appreciated.
For the first part, the answer is in a comment.
For the second part, Let $\lambda\in\sigma(x)$. Then $x-\lambda 1$ is not invertible. By proposition 3.2.4, There is a maximal ideal $\mathcal{I}$ such that $x-\lambda 1\in \mathcal{I}$. By lemma 3.2.8(a) there is a $\phi\in\hat{A}$ such that $\mathcal{I}=\ker\phi$. Then $\phi(x-\lambda 1) = 0$ so $\hat{x}(\phi) =\phi(x) = \lambda\phi(1) =\lambda$, therefore $\lambda\in\hat{x}(\hat{A})$.