I come across this proof: Suppose $X$ is a manifold and let $\mathcal{O}_X$ be the sheaf of real valued functions. We check that $(X,\mathcal{O}_X)$ forms a locally ringed space. Given $x \in X$, let $m_{X,x}$ be the ideal of $\mathcal{O}_{X,x}$ consisting of germs of functions taking the value $0$ at $x$. This is clearly an ideal, and the quotient $\mathcal{O}_{X,x}/m_{X,x}$ is certainly contained in $\mathbb{R}$. Since $X$ is a manifold, the quotient is nonzero, so $m_{X,x}$ is indeed a maximal ideal of $\mathcal{O}_{X,x}$. To check that it is the unique maximal ideal, it suffices to check that any $f\in \mathcal{O}_{X,x}$ not contained in $m_{X,x}$ is a unit in $\mathcal{O}_{X,x}$. For such an $f$, $f(x)$ is some nonzero real number, so we can find an open subinterval $I\subset\mathbb{R}$ such that $f(x)$ belongs to $I$ but $0$ does not. Represent $f$ by a continuous function on some open subset $U$ of $X$ containing $x$, which I’ll also call $f$. The key point is that by continuity, $V = f^{-1}(I)$ is again an open subset of $X$ containing $x$, and $f$ takes nonzero values everywhere on $V$. Hence there exists a multiplicative inverse $g$ of f on $V$, which is necessarily continuous.
I am confused why quotient $\mathcal{O}_{X,x}/m_{X,x}$ is certainly contained in $\mathbb{R}$?
Also, I can't see why since $X$ is a manifold, the quotient is nonzero, $m_{X,x}$ is indeed a maximal ideal of $\mathcal{O}_{X,x}$? How is that related to the manifold and why the quotient is nonzero implies that the the ideal is maximal?
We have a well-defined homomorphism $\mathcal{O}_{X,p} \to \mathbb{R}$ given by taking a germ $(f, U \ni p)$ to the real number $f(p)$. This is well-defined on germs since if two functions $f,g$ agree on a neighborhood of $p$, then they have the same value at $p$.
The kernel of this homomorphism is exactly the set of germs with $f(p) = 0$, so by the universal property of the quotient, we have a well-defined injective homomorphism $\mathcal{O}_{x,p}/m_{X,p} \to \mathbb{R}$ given by taking $[f]$ (the equivalence class of a germ in the quotient) to its value at $p$. Any other germ $g$ in the equivalence class has the same value at $p$, since their difference is in $m_{X,p}$ and so $f(p) = g(p)$.
The image of this homomorphism is all values achieved by continuous functions at $p$. Since the constant functions $f(x) = c$ are always continuous on a manifold, this homomorphism is surjective and thus an isomorphism, $\mathcal{O}_{x,p}/m_{X,p} \cong \mathbb{R}$. So the ideal is maximal since its quotient is a field.