I have done part $(a)$ by defining a map from $C(X,\mathbb R) \to \mathbb R $ as $\phi (f)=f(x) $ and got the $M_x$ as kernel of homomorphism and got the answer.
But I am unable to solve for part $(b).$
Any Hint will be appreciated.
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It is sufficient to show that $M\subset M_x$ for some $x\in X$. Now assume this is not the case, i.e. for all $x\in X$, there exists a function $f_x\in M$ but $f_x\notin M_x$. Then there exists an open $U_x\in X$ such that $\forall y\in U_x$, $|f_x(y)|>0$ and $x\in U_x$.
The sets $\left\{U_x\right\}_{x\in X}$ form an open cover of $X$, by compactness, there is a finite subcover, say $\left\{U_{x_1},\dots , U_{x_n}\right\}$.
Define $g(x)=\sum_{i=1}^nf_{x_i}(x)^2$, then $g:X\rightarrow \mathbb{R}$ is a continuous function such that $g>0$ on $X$ and $g\in M$. Clearly $g$ has an inverse, thus the constant one function is in $M$. But that's a contradiction to $M$ being maximal.
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Mathematician 42
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@Mathematician 42 Do you use that X is Hausdorff in this answer? It does not seem necessary. – J. De Ro Jan 05 '20 at 00:04