Commutative unital Banach algebra not isomorphic to $C(X)$ for some compact Hausdorff space $X$

Here is an argument. Assume that $\pi:\mathcal A\to C(X)$ is a unital Banach algebra isomorphism. Let $f\in \mathcal A$. Put $g=\pi^{-1}(\overline{\pi(f)})$ (that is, map $f$ to $C(X)$, conjugate it, and come back). Now $$ \pi(gf)=\pi(g)\pi(f)=|\pi(f)|^2\geq0. $$ Then, for any $r>0$, $\pi(gf+r+is)=|\pi(f)|^2+r+is$ takes values at distance $r$ or more from $0$, so $gf+r+is$ is invertible. This says that $-r+is$ (we can write a plus since $s$ was arbitrary) is not in the image of $gf$. In other words, $\operatorname{Re}(gf)\geq0$. You can see proof here (applied to $-gf$, and it may require a rotation if $gf(0)$ is not real) that then $gf$ is constant. In other words, there exists $c\in\mathbb R$ such that $gf=c1$. Then $$ |\pi(f)|^2=\pi(gf)=\pi(c1)=c1. $$ Now $f$ was arbitrary and $\pi$ is onto, so every $h\in C(X)$ has $|h|^2$ constant. This can only happen if $X$ consists of a single point, and in that case $\mathcal A$ would be one-dimensional, a contradiction.

Every $C^*$ algebra of dimension at least $2$ must have zero divisor but the disc algebra has no zero divisor.

Proof of existence of zero divisor is based on decomposition $x=x^+ -x^-$ for a self adjoint element $x$. This decomposition is regarded as classical decomposition $f=f^+ -f^-$ with $f^+.f^-=0$ for real valued function $f\in C(X)$.