I need to show that $\mathcal{A}$ closed and $\|{\phi(A)}\|=\|{A}\|$ imply that $\phi(\mathcal{A})$ is closed in $C(\Delta(X))$ (I don't think this is necessary but $\mathcal{A}$ is a $C^*$-algebra, $\phi$ is the Gel'fand transform and $\Delta(X)$ the structure space).
Since $\mathcal{A}$ is closed it follows that $A\in\mathcal{A}$ for a sequence $(A_n)_n\subset \mathcal{A}$ s.t. $A_n\to A$. Then the norm relation yields that also $\phi(A_n)\to\phi(A)$ and obviously $\phi(A)\in C(\Delta(X))$. But this doesn't suffices to show the statement right? I need to show that for a sequence $(\phi(A_n))_n$ converging to an element $B$ it holds that $B\in C(\Delta(X))$.
Thanks for your help.
I am not quite sure if I understand your question correctly, so please let me know if this is not what you are looking for.
It sounds to me like what you want to show is that the image of a Banach space under an isometry is closed. In particular, it follows that the image of a $C^*$-algebra under an isometry is closed. Indeed, consider a Banach space $X$ and let $T \colon X \to Y$ be an isometry into some normed space $Y$. Consider a sequence $(Tx_n)_n$ in the image of $T$ converging to some $y \in Y$. Since $T$ is an isometry, it follows that $(x_n)_n$ is Cauchy in $X$ and thus converges to some $x \in X$. Furthermore, by continuity we have $$Tx = \lim_n Tx_n = y.$$ Hence, $y = Tx$ and we conclude that $T$ has closed image.