Topological space maps naturally to max-spectrum of ring of continuous real-valued functions

Statement
Suppose $$X$$ is a topological space. Let $$C(X,\R)$$ denote the ring of continuous real-valued functions on $$X$$. Then, there is a natural injective (set-theoretic) map to the max-spectrum $$\operatorname{Max-Spec}(C(X,\R))$$:

$$X \to \operatorname{Max-Spec}(C(X,\R))$$

given as follows:

$$x \mapsto M_x = \{ f \in C(X,\R) \mid f(x) = 0 \}$$.

Proof
Given: A topological space $$X$$, the ring $$C(X,\R)$$ of continuous real-valued functions on $$X$$. A point $$x \in X$$. $$M_x = \{ f \in C(X,\R) \mid f(x) = 0 \}$$.

To prove: $$M_x$$ is a maximal ideal in $$C(X,\R)$$.

Proof: Consider the map:

$$\operatorname{eval}_x: C(X,\R) \to R$$

given by:

$$\operatorname{eval}_x(f) = f(x)$$.

By the definition of pointwise addition and multiplication, $$\operatorname{eval}_x$$ is a homomorphism from $$C(X,\R)$$ to $$\R$$. Further, since constant functions are continuous, every $$c \in \R$$ can be expressed as $$\operatorname{eval}_x(f)$$ where $$f$$ is the constant function $$c$$. Thus, $$\operatorname{eval}_x$$ is surjective. Thus, $$\operatorname{eval}_x$$ is a surjective homomorphism to a field, so its kernel, which is $$M_x$$, is a maximal ideal.