If $X$ is a compact topological space then $C(X)$ is a C* algebra. I'm not going to attempt to discuss the locally compact case with $C_0(X)$ because usually the definition of that space requires the underlying $X$ to be Hausdorff. $C(X)$ has a Gel'fand representation, as some $C(Y)$ with Y compact and Hausdorff. How does Y relate to X? Is it via some standard process by which one can take a nonHausdorff space and maybe design some strange equivalence relation for which the fact set is $Y$, and Hausdorff?
Let $\delta_t$ be the evaluation at the point $t\in X$. It is easy to check that $s\sim t\iff\delta_s=\delta_t$ is an equivalence relation $\sim$ on $X$. Consider $X/\sim$, with the quotient topology (i.e. the smallest topology that makes the quotient map continuous).
Now consider the map $\phi:C(X/\sim)\to C(X)$ given by $\phi(f)(t)=f([t])$, where $[t]$ is the class of $t$ in the quotient. This map is obviously a $*$-homomorphism. It is one-to-one by construction, because if $\phi(f)=0$, then $f([t])=0$ for all $[t]$, i.e. $f=0$. And it is onto: if $g\in C(X)$, then we define a function $g'\in C(X/\sim)$ by $g'([t])=g(t)$; this of course implies that $\phi(g')=g$. All we need to check is that such $g'$ can be defined; this is exactly to say that $g(t)$ does not depend on the value of $g$, and this holds, since $s\sim t$ if and only if $\delta_s=\delta_t$.
The topology on $X/\sim$ is Hausdorff: given $[s]\ne[t]\in X/\sim$, we have that $\delta_s\ne\delta_t$, so there exists $f\in C(X)$ such that $f(s)\ne f(t)$. The construction of the quotient topology guarantees that $f'$, as in the previous paragraph, is continuous; we get that $f'([s])\ne f'([t])$, and so $[s]$ and $[t]$ can be separated.
So, as C$^*$-algebras, $C(Y)\simeq C(X)\simeq C(X/\sim)$. Since $Y$ and $X/\sim$ are Hausdorff, we deduce that $Y$ is homeomorphic to $X/\sim$.