The Spectrum of C(X) is Hausdorff for X compact, hausdorff. Let $X$ be a compact hausdorff space. Define $C(X) = \{f:X\to \Bbb R, f \text{ continuous}\}$. $C(X)$ is a commutative ring and define $ev_x$ to be the maximal ideal defined as the kernel of the map: $C(X) \to \Bbb R,f \to f(x) $.
I can show that these are all the maximal ideals of $C(X)$, define $\tilde X$ to be the set of all the maximal ideals of $C(X)$ with the following closed sets for $f\in C(X)$:
$$V(f) = \{\mathfrak m \in \tilde X : f\in \mathfrak m\}.$$
We can define a map $\varphi: X \to \tilde X, \varphi(x) = ev_x$. It is easy to show that it is continuous and bijective. It is also easy to show that $\tilde X$ is compact.
Question: I would like to show that $\varphi $ is a homeomorphism.
One way to do this would be to show that $\tilde X$ is hausdorff. This involves showing that given $x,y \in X$, there exist $f,g \in C(X)$ such that $x \not\in V(f), y\not\in V(g)$ and $V(f) \cap V(g) = X$.
Alternatively, one can show that $\varphi$ is a closed map. This reduces to showing that given a closed subset $V\subset X$, one can find a function $f_1,\dots,f_n \in C(X)$ such that $\bigcap_k f^{-1}(0) = V$ or equivalently, $f = \sum_k f_k^2$ vanishes precisely on $V$.
Therefore, we can reduce the second attempt to showing that there exists some $f\in C(X)$ such that $f^{-1}(0) = V$, ie, $f$ vanishes exactly on $V$.
I know that the proof involved using some version of Urysohn's lemma but I have not been able to finish it.
 A: Note that you are giving the Zariski topology (w.r.t. continuous functions) to $X$. This will give the original topology of $X$ if it defines the same closed sets, i.e., if closed sets of $X$ are precisely intersections of zero loci.
This property corresponds to $X$ being a Tychonoff space. Every locally compact Hausdorff space is Tychonoff, so we are done.
More generally, the topology of maximal ideals of an algebra $A$ is Hausdorff if and only if the quotient of $A$ by its Jacobson radical is a Gelfand ring. 
In the language of spaces, it is roughly equivalent to asking the existence of "many" pairs of closed sets $E,F$ whose union is everything. 
As a reality check, Tychonoff spaces are the ones for which continuous functions separate closed sets by points. So given two distinct points, you can build two functions $e,f$ using appropriately chosen $E,F$, and they will satisfy $ef=0$. Then $V(e)^c,V(f)^c$ is a pair of open sets satisfying the Hausdorff property.
A: I know some time has passed since this question was asked and answered, but at least for me vap's answer uses some advanced tools which I am not familiar with yet.
Therefore, for the sake of other topology newbies like myself I thought it would be nice to post a proof which uses some more elementary tools:
So I will prove that $\varphi$ is a homeomorphism using the first option suggested, that is I will show $\tilde X$ is hausdorff.
As the OP suggested it can be shown that each ideal $m \in \tilde X$ is of the form $m_y=\{f\in C(X) | f(y)=0\}$ for some $y\in X$.
So let $m_{x_1}$, $m_{x_2}$ be two different ideals in $\tilde X$. Since $m_{x_1}\ne m_{x_2} $ we conclude $x_1 \ne x_2$, and since $X$ is hausdorff we get that there exist $U_{x_1},U_{x_2} \subseteq X$ open and disjoint neighborhoods of $x_1,x_2$ respectively.
So $X\setminus U_{x_1}$ and $X\setminus U_{x_2}$ are closed in $X$ and $X\setminus U_{x_1} \cap \{x_1\}=\varnothing$ and $X\setminus U_{x_2} \cap \{x_2\} = \varnothing$. Also, note that $X$ being hausdorff means that every singleton is a closed set.
Finally, $X$ being compact and hausdorff implies it is normal. From Urysohn's lemma we get continuous $h_1,h_2:X\to [0,1]$ such that:
$$h_1(X\setminus U_{x_1})=\{0\}, h_1(x_1)=1\ne 0$$
$$h_2(X\setminus U_{x_2})=\{0\}, h_2(x_2)=1\ne 0$$
So we have $h_1(x_1)\ne 0 \implies m_{x_1}\in V(h_1)^c$ and $h_2(x_2)\ne 0 \implies m_{x_2}\in V(h_2)^c$, where as complements of closed sets both are open neighborhoods of $m_{x_1}$ and $m_{x_2}$ in $\tilde X$ respectively.
Lastly, we show $V(h_1)^c \cap V(h_2)^c = \varnothing$. Suppose towards contradiction that there exists $m_y \in V(h_1)^c \cap V(h_2)^c$ for some $y\in X$, this means that:
$$m_y\in V(h_1)^c \implies h_1(y) \ne 0 \implies y \notin X\setminus U_{x_1} \implies y\in U_{x_1}$$
$$m_y\in V(h_2)^c \implies h_2(y) \ne 0 \implies y \notin X\setminus U_{x_2} \implies y\in U_{x_2}$$ 
But $U_{x_1}$ and $U_{x_2}$ are disjoint, so we get a contradiction.
Thus for arbitrary two different elements $m_{x_1},m_{x_2}\in \tilde X$ we found open neighborhoods $V(h_1)^c,V(h_2)^c$ in $\tilde X$ such that $m_{x_1} \in V(h_1)^c$ and $m_{x_2} \in V(h_2)^c$ and $V(h_1)^c \cap V(h_2)^c = \varnothing$. Hence $\tilde X$ is hausdorff.
Just so the answer would be complete, as one can show that $\varphi$ is continuous and bijective, it just remains to notice that a continuous bijection from a compact space to a hausdorff space is a homeomorphism.
Hopes this helps someone.
