I know this question has been asked before, but I think I'm very close to a new solution and wanted to know if it is a viable approach.
Let $C(X)$ be the ring of continuous functions $X \rightarrow \mathbb{C}$ where $X$ is compact Hausdorff, and let $\mathfrak M$ be a maximal ideal. We want to show that $\mathfrak M = I_{x_0} = \{ f \in C(X) : f(x_0) = 0\}$ for some $x_0$.
For each nonempty closed set $A$, we define the ideal $I_A = \{ f \in C(X) : f(x) = 0, \forall x \in A\}$. Clearly $A \subset B$ implies $I_B \subset I_A$ (I also need to show that $I_B \subseteq I_A$ implies $A \subseteq B$, which I haven't done yet). We let $\mathcal S$ be the set of ideals $I_A$ which are contained in $\mathfrak M$.
An ascending chain $I_{A_i} \in \mathcal S$ corresponds to a descending chain of closed sets $A_i$, whose intersection is nonempty because $X$ is compact. Thus the union $J$ of the chain $I_{A_i}$ is an ideal which is contained in $I_{\bigcap A_i}$ (now we have to show that $I_{\bigcap A_i} \subseteq \mathfrak M$ to show that it is an upper bound for the chain, I haven't been able to do this).
So by Zorn's Lemma, $\mathcal S$ has a maximal element $I_D$ for some nonempty closed set $D$. Now, I just want to show that $D$ is a singleton set $\{x_0\}$; then $I_D$ is a maximal ideal contained in $\mathfrak M$, whence $I_D = \mathfrak M$. That will finish the proof.
Edit: Assuming I can fix the holes, I finished the problem (answered below).
Current problems with the proof:
(i) Need to show that $I_B \subseteq I_A$ implies $A \subseteq B$.
(ii) Need to finish the Zorn's lemma argument by showing that $I_{\bigcap\limits_i A_i} \subseteq \mathfrak M$.
Second edit: (i) follows from the Urysohn lemma. A compact Hausdorff space is normal. If $x \in A$, but not in $B$, then $\{x\}$ and $B$ are disjoint closed sets, so by Urysohn there exists a continuous function $f: X \rightarrow [0,1] \subseteq \mathbb{C}$ for which $f(x) = 1$ and $f(b) = 0$ for all $b \in B$. Thus $f \in I_B$, but not in $I_A$. So there is only one hole in the proof left.
Third edit: (ii) also follows from the Urysohn lemma. See the edited answer for details.
Best Answer
It's even easier. Take a strict ideal $I$ and show, as you intended to, that $I\subseteq I_x$ for some $x\in X$. For this, suppose the contrary : then for all $x\in X$ you can find an $f_x \in I$ such that $f_x \not\in I_x$ i.e. $f_x (x) \not= 0$. Since $f_x$ is continuous there is a open neighbour hood of $x$ in $X$ such that $f_x(y)\not=0$ for all $y\in U_x$. Now $(U_x)_{x\in X}$ is an open cover of the compact space $X$, and has therefore a finite subcover : we can find $x_1,\ldots,x_n \in X$ such that $X = \cup_{i=1}^n U_{x_i}$. Consider the function $f = \sum_{i=1}^n |f_{x_i}|^2 = \sum_{i=1}^n \overline{f_{x_i}} f_{x_i}$. This $f$ is obviously in $I$ as each $f_{x_i}$ is an as $I$ is an ideal. But by construction $f > 0$. Indeed : let $x\in X$. That $x$ is in some $U_{x_i}$ and then $f(x)\geq |f_i (x)|^2 > 0$ by definition of the $U_x$'s. Now, this function is invertible, so that $I$ contains an invertible element, so that $I$ is not strict. Absurd.