The following result are true if we assume full axiom of choice:
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A. If $X$ is a compact Hausdorff space, then every maximal ideal of the ring $C(X)$ has the form $A_p=\{f\in C(X); f(p)=0\}$.
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B. If $X$ is a compact Hausdorff space, then every ideal of the ring $C(X)$ is contained in an ideal of the form $A_p=\{f\in C(X); f(p)=0\}$.
I wonder how much choice is needed. To be precise, do we get a statement equivalent to some known form of AC if we assume validity of A/B for every compact space, for every compact metric space, for every complete totally bounded metric space or for the case $X=[0,1]$?
I've tried to answer A at least partially in my answer here.
If I did not make a mistake there, I've shown that in ZF the claim A holds for complete totally bounded metric space, B holds for compact spaces. I've also gathered a few relevant references in that answer. According to those references validity of A of every compact regular space is equivalent to ultrafilter theorem. According to the same book, in ZF it can be shown that A holds if and only if $X$ the form $[0,1]^I$ and B holds if and only of $X$ is compact.
As I am not experienced with working in ZF (without AC), I'll be glad if you check my work there and point out any mistakes and add any additional references/proofs/insights.
This question is also related, but not identical: https://math.stackexchange.com/questions/97603/realizing-a-homomorphism-mathcalcx-to-mathbbr-as-an-evaluation
Best Answer
EDIT : The following does not answer the question, because it uses Urysohn's lemma, as pointed out in the comments. However, the only choice needed is, I think, countable dependent choice. It also gives a proof in ZF of both A and B simultaneously, in the case of compact metric space.
First, let's show that $A_p$ is a maximal ideal. Let $J$ by an ideal of $C(X)$ such that $A_p \subsetneq J$. Let's show that necessarily $J=C(X)$. Let $g \in J$, $g \notin A_p$. So $g(p) \neq 0$, and we can write $$1 = \frac{g(p)-g}{g(p)} + \frac{g}{g(p)}$$ The first term vanishes at $p$, so is in $A_p$, and the second term belongs to $J$. It follows that the function $1 \in J$, and thus $J=C(X)$.
Now, let $M$ be a maximal ideal of $C(X)$. Suppose that for all $p \in X$, there exists $g_p \in M$ such that $g_p(p) \neq 0$. By continuity, $g_p \neq 0$ on some neighborhood $V_p$ of $p$. By compacity of $X$, there exists $p_1, p_2, \dots, p_n$ such that $$X=V_{p_1} \cup V_{p_2} \cup \dots \cup V_{p_n}.$$ Let $h_1, h_2, \dots, h_n$ be a partition of unity subordinate to $V_{p_1}, \dots, V_{p_n}$ (This follows from Urysohn's lemma). Now write $$1= h_1 + \cdots + h_n = \frac{h_1}{g_{p_1}}g_{p_1} + \frac{h_2}{g_{p_2}}g_{p_2} + \cdots + \frac{h_n}{g_{p_n}}g_{p_n}.$$ The functions $h_j/g_{p_j}$ are continuous in $X$, since the support of each $h_j$ is contained in $V_{p_j}$, and $g_{p_j} \neq 0$ in $V_{p_j}$. It follows that $1 \in M$, a contradiction.
We have shown that there exists a $p \in X$ such that for all $g \in M$, $g(p)=0$, i.e. $M \subseteq A_p$. By maximality of $M$, $M=A_p$.
The second part of the proof also shows that $B$ holds.