Show that there exists a $t \in [0, 1]$ such that $\alpha(f) =f(t)$ for all $f \in C_\mathbb{R}[0, 1]$. 
Suppose $\alpha$ is a nonzero multiplicative linear functional on $C_\mathbb{R}[0, 1]$. Show that there exists a $t \in [0, 1]$ such that $\alpha(f) =f(t)$ for all $f \in C_\mathbb{R}[0, 1]$. 

I approached this by contradiction. I defined $g_t(x) = f_t(x) - \alpha(f_t)$ such that when I apply $\alpha$ to it, I get $0$ with $g_t(t) \ne 0$. I take a finite subcover of $[0,1]$ where each open set is centered around a $t$. I then define $G(x) = \sum\limits_{j = 1}^\infty (g_{t_j}(x))^2$ (note that $G \ne 0$ for all $x \in [0,1]$). I want to apply $\alpha$ to $G$ to get $0$, and then get a contradiction with $1 = \alpha(1) = \alpha(G \frac{1}{G}) = \alpha(G) \alpha(\frac{1}{G})$, but I don't know if I can apply $\alpha$ to an infinite sum. Any help would be appreciated.
 A: I think your argument is good since you actually don't have an infinite sum:
After defining $g_t$ for each $t\in [0,1]$ there is an open neighbourhood $U_t$  of $t$ such that $g_t$ is never zero on $U_t$. Then you choose a $\textbf{finite}$ subcover $U_{t_1},...,U_{t_n}$, define $G=\sum_{i=1}^ng_{t_i}^2$ and conlude as  suggested.
A: Nice! 
Here's a more direct approach. 


*

*$\alpha (f) = \alpha(1 f ) = \alpha(1)\alpha(f)$.  Therefore 


$$ (*)\quad \alpha(1) =1,$$ 
because $\alpha$ is not identically zero. 


*

*If $f\ge 0$
$$ (**) \quad 0 \le \alpha (\sqrt{f})^2 = \alpha(f).$$ 


*

*Since $0\le \|f\|_\infty-f$, it follows that 


$$ (***) \quad  |\alpha(f)| \le \alpha (\|f\|_\infty) = \|f\|_\infty.$$ 


*

*By  $(**)$, $\alpha(x) \ge 0$, and then  by $(***)$, $\alpha (x) \in [0,1]$. Therefore by the intermediate value theorem there exists $x_0$ satisfying 


$$ \alpha (x) = x_0.$$ 
This, along with $(*)$, linearity and multiplicative property imply 
$$\alpha (p) = p(x_0),$$ 
whenever $p$ is a polynomial. 


*

*As polynomials are dense in $C[0,1]$, it follows that for every $f$, there exists a polynomial $P_n$ such that $\| f-P_n\|_\infty \le \frac1n$. Thus


$$ |\alpha(f) - f(x_0) | = \underset{\le \frac {1}{n}}{\underbrace{ |\alpha (f) - \alpha(P_n) |}} + \underset{=0}{\underbrace{|\alpha(P_n) - P_n (x_0)|}} + \underset{\le \frac 1n}{\underbrace{| P_n (x_0) - f(x_0)|}},$$
with first inequality on RHS is due to $(***)$. This completes the proof. 
