What if every function in $C(X)$ has finite spectrum? Suppose that $X$ is a compact Hausdorf space, and that every continuous function on $X$ has finite range.  How do I conclude that $X$ is a finite set, hence with discrete topology?  So far, I have managed to use Urysohn to show that the closed maximal connected component decomposition of $X$ (as opposed to the path component decomposition) is one where every component is a 1 point set.
This question arose from trying to solve the question that every C* algebra with all normal elements having finite spectra are finite dimensional.  A more direct solution to this general question would also suffice, because actually implies the special case I've reduced it to.  By taking a unitization, and then a MASA in the unitization, one gets to the case I've gotten to.
Also, a topological question that is in the same setup but handles a related operator algebras question is this: if a net $u_i$ and $u$ in $X$ has the property that for all continuous $f \in C(X)$ one has that $f(u_i)$ converges to $f(u)$ then must $u_i$ converge to $u$ in topology?  This is for showing that the gelfand representation of a $C(X)$ space is itself.  I'd like this fact for $X$ locally compact Hausdorf, actually.
 A: For your first question, let $x\in X$ be arbitrary, and let $U$ be any open set containing $x$. Let $f:X\to[0,1]$ be a continuous function such that $f(x)=0$ and $f(y)=1$ for all $y\in X\setminus U$, and let $\alpha$ be the smallest non-zero element of the range of $f$. Then $V=\left\{y\in X:f(y)<\frac{\alpha}2\right\}$ is a clopen neighborhood of $x$ contained in $U$. Thus, $X$ has a clopen base, i.e., is zero-dimensional. This means that if $p$ is a non-isolated point of $X$, we can construct a sequence $\langle U_n:n\in\Bbb N\rangle$ of clopen neighborhoods of $p$ such that $U_0=X$, $U_{n+1}\subseteq U_n$ for each $n\in\Bbb N$, and for each $n\in\Bbb N$ there are distinct points $x_n,y_n\in U_n\setminus U_{n+1}$. Let $K=\bigcap_{n\in\Bbb N}U_n$.
For each $n\in\Bbb N$ there is a continuous function $$f_n:U_n\setminus U_{n+1}\to\left[
\frac1{2^{n+1}},\frac1{2^n}\right]$$ such that $f(x_n)=\dfrac1{2^n}$ and $f(y_n)=\dfrac1{2^{n+1}}$. Define
$$f:X\to[0,1]:x\mapsto\begin{cases}
f_n(x),&\text{if }x\in U_n\setminus U_{n+1}\\
0,&\text{if }x\in K\;.
\end{cases}$$
(In other words, $f=\sum_{n\in N}f_n$ on $X\setminus K$, and $f[K]=\{0\}$.) Clearly $f$ is continuous on $U\setminus K$, since each part of its definition has clopen domain, and the construction also ensures that $f$ is continuous on $K$. But the range of $f$ includes $2^{-n}$ for every $n\in\Bbb N$, so it’s infinite. Thus, $X$ cannot have a non-isolated point and must therefore be finite.
Added: For the second question, suppose that the net $\nu=\langle x_i:i\in I\rangle\not\to x$, where $\langle I,\le\rangle $ is a directed set. Then $x$ has an open neighborhood $U$ such that $\nu$ is frequently in $X\setminus U$. Let $f:X\to[0,1]$ be a continuous function such that $f(x)=0$ and $f(y)=1$ for $y\in X\setminus U$, and suppose that $\langle f(x_i):i\in I\rangle\to 0=f(x)$. Then there is an $i_0\in I$ such that $f(x_i)<\frac12$ for all $i\ge i_0$. Clearly $\left\{y\in X:f(y)<\frac12\right\}\subseteq U$, so there is an $i\ge i_0$ such that $x_i\notin U$ and hence $f(x_i)=1$; this contradiction shows that $\langle f(x_i):i\in I\rangle\not\to f(x)$.
In other words, if $\langle f(x_i):i\in I\rangle\to f(x)$ for every $f\in C(X,\Bbb R)$, then $\langle x_i:i\in I\rangle\to x$ in $X$.
A: For the first question. suppose X has an infinite sequence of point $x_0, ...,x_n,...$.
for each $i$ considere a function $f$ such that $f_i(x_0)= 0$,...,$f_i(x_{i-1})=0$, $f_i(x_i)=1$, and $\Vert f_i \Vert \leqslant 1$
(Uryshon can do that, just consider the function on {x_0...x_i} which is a discret space and then use the extension properties...) 
Let  $ g = \sum \frac{f_i}{2^i}$.
then $\frac{1}{2^{i-1}} \geqslant g(x_i)\geqslant \frac{1}{2^i}$
ensure you that $g$ takes an infinite number of value.
A: For your second question, if X is compact, you can extract a converging subnet from (u_i) and let u' be its limit... you have f(u)=f(u') for every continuous f, and hence u=u'. If X is just locally compact, then consider its one point 'alexandrov' compactification. (and if he is just regular, then the Stone Cech compactification will do the trick).
