Riesz Representation and Ring Homomorphism Let $X$ be a compact metric space. Show that any ring homomorphism $C(X)\to \mathbb{R}$ is an evaluation $f \to f(x_0)$ for some point $x_0 \in X$.
$\textbf{Question}$:
have the sketch of the proof as outlined below. but Not sure of from the step 3 and also have problem with the final step.
$\textbf{Sketch of the Proof}$:
1) Show that such a homomorphism is a positive linear functional. 
2) Apply the Riesz representation theorem to show that it is
given by integration with respect to some measure.
3) Use the homomorphism property to argue that if U, V are disjoint open sets at most one can have positive measure. 
4) Deduce that the measure in question is a Dirac delta measure at some point.
$\textbf{My attempt}$:
let $\Lambda :C(X)\to \mathbb{R}$, then by assumption it a ring homomorphism and have the following properties for $f,g \in C(X)$; we have that $(i)$ $\Lambda(f+g) = \Lambda(f)+\Lambda(g)$ ,$(ii)$ $\Lambda(f.g) = \Lambda(f).\Lambda(g)$ , and $(iii)$ $\Lambda(1_{X}) = 1$ or $\Lambda(0_X) = 0$ 
$\textit{Step 1} :$ I need to show that if $f\geq0$ then $\Lambda(f)\geq0$. 
let $f\geq0$ be any continuous function, we can approximate that by simple functions  $ f = \lim_{n\to \infty}S_n$ . Applying the operator $\Lambda (f) =\lim_{n\to \infty}\Lambda (S_n) $ ;(I need to prove that $\Lambda$ is continuous.)   So it suffices to show that 
$A)$ $\Lambda (S)\geq0 $ for any simple function $S=\sum_{i=1}^\infty a_i\mathcal{X_{A_{i}}}$ , $a_i\geq 0$
$B)$ $\Lambda (\mathcal{X_{A}})\geq0 $ 
I will show $B$ then $A$.
let $\{f_n\}_{n=1}^\infty$ be a sequence of continuous functions that we approximate the characteristic function (as it is not continuous) by them.  So I start with a characteristic function $\lim_{n\to \infty} f_n^A = \mathcal{X_A}$. 
Clearly characteristic function satisfies the above $(i)-(iii)$ and clearly since it takes zero or one it is positive.
To show A we can write that by $(i)$ , $\Lambda(S) =\sum_{i=1}^\infty a_i\Lambda(\mathcal{X_{A_{i}})} \geq 0$ as $a_i\geq0$
$\textit{Step 2} :$ Then since $\Lambda$ is a positive linear operator by Riesz representation we have that $\exists$ measure $\nu$ that we write the functional as $\Lambda(f) = \int_X fd\nu $
$\textit{Step 3} :$ Let $U,V \subset X$ be disjoint. Then we can write for any two characteristic functions (which we'd assume that are the limit of continuous functions) 
$\Lambda (\mathcal{X}_U).\Lambda (\mathcal{X}_V) = 0 $ so at most one of them can be positive. Also we have that 
$\Lambda (\mathcal{X}_U) =  \int_X \mathcal{X}_U d\nu =\int_U d\nu = \nu(U)$
so $\nu(U).\nu(V) =0 $ which implies that at most one of them can be positive.
$\textit{Step 4} :$ 
 A: About your specific solution: as was pointed out in the comments you exchange $\lim$ and $\Lambda$ without justification, which is problematic as it's not clear that they can be exchanged (especially if $\lim$ is to be taken in the sense of simple convergence of functions). 
Here are some hints on how to solve the problem (not about your specific solution):
Hints : Use square roots to prove positivity. 
Use positivity to prove contintuity, at least on constant functions : deduce from that that $\Lambda$ is actually an $\mathbb R$-algebra morphism. In particular, unless $X=\emptyset$, $\Lambda$ is surjective and so its kernel is a maximal ideal of $C(X)$. 
Study maximal ideals of $C(X)$ as follows : 
Let $m$ be such an ideal, 
i) Assume for all $x\in X$ there is $f\in m$ with $f(x) \neq 0$. Use compactness to find $f\in m$ that never vanishes. 
ii) Deduce (using that nonvanishing implies invertible) that there is some $x_0$ with $f(x_0) = 0$ for all $f\in m$. 
iii) Prove that $I=\{f \in C(X)\mid f(x_0) = 0\}$ is a proper ideal. In particular $m\subset I \implies m= I$ 
Once you have the shape of maximal ideals you can deduce that $f$ is of the required shape. 
You can bypass the first steps if you can find some other way to prove surjectivity some other way (I can think of at least one way to do so, although in essence it's the same thing)
A: Since $X$ is compact, for every $n$, there exists a finite number of closed balls $B(x^n_1,{1\over n}),...,B(x^n_{i_n},{1\over n})$ of $X$ which cover $X$, since $\nu(\cup B(x^n_{i_j},{1\over n}))\neq 0$, we deduce that there exists a ball $B_n=B(x^n_{i_j},{1\over n}))$ such that $\nu(B_n)\neq 0$. 
Consider $C=\cap_nB_n$. Suppose that $C$ is empty, there existsa finite number of $B_n$,
$B_{n_1},...,B_{n_p}$ whose intersection is empty. (Baire). This implies (recursively) that there exists $B_{n_i}, B_{n_j}$ whose intersection is empty contradiction with the previous step.
$C$ contains one element $x$, and if $U$ is an open subset which does not contains $x$, for every element $y$ of $U$, there exists $B_n,$ such that $x\in B_n$ and $B(y,{1\over n})\cap B_n$ is empty, and $B(y,{1\over n})\subset U$ this implies that $\nu(B(y,{1\over n})=0$ and $\nu(U)=0$. This shows that the support of $\nu$ is $x$ and $\nu=\delta_x$.
