Let $X$ be a compact Hausdorff space and $\Omega (C(X))$ the space of characters on $C(X)$. I am showing that the map $x \mapsto e_x$ where $e_x$ is evaluation at $x$ is surjective (I already showed it's injective). But I just can't understand one step in the proof I am reading. Let me give the proof here:
Let $\tau \in \Omega (C(X))$ be a character. Then $K = \operatorname{ker}{\tau}$ is a maximal ideal in $C(X)$ and therefore proper and closed. Using Urysohn's lemma it is easy to see that $K$ separates points. Since $K$ is proper and contains $1$ it cannot vanish nowhere hence there must exist $x \in X$ such that for all $f \in K$ we have that $f(x) = 0$.
So far so good. Now here comes the step I don't understand:
Hence $(f-\tau(f))(x) = 0$ so $f(x) = \tau(f) \color{red}{\text{ for all }}$ $f\in C(X)$.
It is clear to me that if $f \in K$ then $\tau(f) = 0$ and by how $x$ was chosen also $f(x) = 0$ and therefore the above claim holds for all $f \in K$. But why does it hold for all $f \in C(X)$?
You have $0=e_{x}(f)=\tau(f)$ $\forall f\in K$
So, $e_{x}$ and $\tau$ are two linear functionals that are equals in the hyperplane $K$. Also, $e_{x}(1)=\tau(1)=1$ since $\tau$ is multiplicative. Then, $e_{x}=\tau$