I am new to $C^{*}$-algebras and, although I know there are a lot of references in which the functional calculus is proved, I was trying to prove it myself with my own reasoning and words. I have already proved Gelfand Isomorphism Theorem, so it can be used in the proof.
So, let $\mathcal{A}$ be a $C^{*}$-algebra with unit and $a\in \mathcal{A}$ be normal. I want to prove that $C^{*}(a) \cong C(\sigma(a))$, where $C^{*}(a)$ is the unital abelian $C^{*}$-algebra generated by $a$, $\sigma(a)$ is the spectrum of $a$ relative to $\mathcal{A}$ and $\sigma_{C^{*}(a)}(a)$ is the spectrum of $a$ relative to $C^{*}(a)$, which is the same as $\sigma(a)$.
Attempt of Proof: By Gelfand Isomorphism Theorem, I know that $C^{*}(a) \cong C(\widehat{C^{*}(a)})$. The mapping $\hat{a}: \widehat{C^{*}(a)}\to \mathbb{C}$ is injective because every character on $C^{*}(a)$ is completely determined by its values on the generating element $a$. Moreover: $$\operatorname{Im}(\hat{a}) = \{\hat{a}(\varphi): \varphi \in \widehat{C^{*}(a)}\} = \{\varphi(a): \varphi \in \widehat{C^{*}(a)}\} = \sigma_{C^{*}(a)}(a) = \sigma(a) $$ and it follows that $\hat{a}: \widehat{C^{*}(a)} \to \sigma(a)$ is a homomorphism, so $\widehat{C^{*}(a)} \cong \sigma(a)$. Therefore, if $\hat{b} \in C(\widehat{C^{*}(a)})$ then $b\circ \hat{a}^{*} \in C(\sigma(a))$ is an algebra $*$-isomorphism. Thus, $C^{*}(a) \cong C(\sigma(a))$ is an algebra $*$-isomorphism and, in particular, $a \mapsto \hat{a}\circ \hat{a}^{*} = \operatorname{Id}$ is the identity map from $\sigma(a)$ to $\mathbb{C}$.
Is my proof correct?
Your argument is correct, but some clarifications are needed.
You argued that $\hat a$ is bijective. You want to conclude that it is a homeomorphism (you have a typo in there). This requires you to show that it is bicontinuous, which in turn makes look at the topologies involved. The crucial point is to consider the weak$^*$-topology on $\widehat{C^*(a)}$, which makes it compact. The continuity of $\hat a$ now follows from the definition of the weak$^*$-topology, and the continuity of $(\hat a)^{-1}$ follows from the compactness of $\widehat{C^*(a)}$.