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.
A key part of the story here is that a $C^*$-algebra whose self-adjoint elements all have finite spectrum must be finite-dimensional. Martin Argerami gave a reference which implies this result by way of several more general results in the context of Banach algebras. However, I decided to write up an account specifically for the $C^*$-algebra case in order to take advantage of the simplifications this context has to offer.
Claim 1: Suppose $A$ is a unital $C^*$-algebra all of whose self-adjoint elements have one point spectrum. Then $A$ is one-dimensional.
Sketch: Use functional calculus to show every self-adjoint element is a multiple of $1_A$, then use the fact that every $C^*$-algebra is the span of its self-adjoint part.
Claim 2: Let $A$ be a $C^*$-algebra in which every self-adjoint element has a finite spectrum and let $p$ be a nonzero projection in $A$ such that the "corner algebra" $pAp$ is not one dimensional. Then, one can write $p=p_1+p_2$ where $p_1$ and $p_2$ are nonzero, orthogonal projections.
Sketch: Working inside $pAp$, whose unit is $p$, the previous claim gives a self-adjoint element whose spectrum is a finite set with at least two elements. Now use functional calculus.
Claim 3: Let $A$ be unital $C^*$-algebra all of whose self-adjoint elements have finite spectrum. Then there exist nonzero, pairwise orthogonal projections $p_1,\ldots,p_n$ summing to $1_A$ such that each of the corner algebras $p_iAp_i$ is one-dimensional.
Sketch: If $A$ is not one-dimensional, subdivide $1_A$ into two projections. If either of the new projections does not determine a one-dimensional corner, subdivide again. This process must eventually terminate and give a decomposition of the desired type. Otherwise, we would have an infinite collection of pairwise orthogonal projections in $A$ which could then be used to embed a copy of $c_0(\mathbb{N})$ into $A$ leading to self-adjoint elements of infinite spectrum, contrary to assumption.
Claim 4: Let $p$ and $q$ be nonzero, orthogonal projections in a $C^*$-algebra $A$ satisfying that $pAp$ and $qAq$ are one-dimensional. Then $qAp$ and $pAq$ are either both one-dimensional or both zero.
Proof: Since $(qAp)^* = pAq$, it suffices to look at $qAp$. Fix any nonzero $a \in qAp$. By the $C^*$-identity, $a^*a \in pAp$ and $aa^* \in qAq$ are nonzero. Since $pAp$ is one-dimensional, up to rescaling $a$, we may assume $a^*a=p$. But then $a$ is a partial isometry, so $aa^*$ is also a projection and, being a nonzero projection in $qAq$, is equal to $q$. The identities $a^*a=p$ and $aa^*=q$ show that $x \mapsto ax : pAp \to qAp$ and $x \mapsto a^* x : qAp \to pAp$ are mutually inverse bijections, and the result follows.
Claim 5: Let $A$ be a unital $C^*$-algebra and let $p_1,\ldots,p_n$ be (necessarily orthogonal) projections satisfying $1_A=p_1 + \ldots + p_n$. Then, $A$ is the internal direct sum of the spaces $p_i A p_j$ where $1 \leq i,j \leq n$.
Basically, each element of $A$ can be thought of as a "matrix of operators" with respect to this partition of the identity.
Final Claim: Let $A$ be a $C^*$-algebra all of whose self-adjoint elements have finite spectrum. Then, $A$ is finite-dimensional.
Proof: First unitize $A$ if necessary (which does not affect the spectra of any of its elements). Then combine Claims 3, 4 and 5.
Best Answer
The characters are linear maps $A\to\mathbb C$. Being $*$-homomorphisms they are positive (since $\phi(a^*a)=|\phi(a)|^2\geq0$), so they are in $A^*$. Therefore one considers the weak$^*$ topology given by $A$.
Also from being $*$-homeomorphisms, you get that $\|\phi\|=1$ for all $\phi$. So they live in the closed unit ball of $A^*$, which is weak$^*$-compact by Banach-Alaoglu. Thus the closure of the set of characters is compact.
When $A$ is unital, the set of characters is closed; so compact. This is easy to see: a limit of $*$-homomorphisms is a $*$-homomorphism. When $A$ is not unital, the set is not closed because $0$ is in the closure.