Maximal ideal of the ring of all real valued continuous functions on $[0,1]$ 
Let $R$ be the ring of all real valued  continuous functions on $[0,1]$ and $M$ be a maximal ideal of $R$. Then how to prove that $\exists c \in [0,1] $ such that $M:=\{f \in R:f(c)=0\}$ ?    

 A: Note that a function in $R$ is invertible iff it is nowhere zero.
Outline of proof:
First, show that if every point $p$ of $[0,1]$ has a function $f$ in $M$ which is nonzero at $p$ then $M$ contains a nowhere zero function, and hence would be all of $R$.  
Second, show that if an ideal $M$ of $R$ has more than one point at which every function in $M$ vanishes, then $M$ is not maximal.
The second part is easy.  The first part is a bit more involved, so I've written all of the details below in case you are stuck proving it on your own. 

 Suppose every point $p\in[0,1]$ has a function which is nonzero at $p$, and hence by continuity, in a (connected) neighborhood of $p$.  Take one such function and neighborhood for each point of $[0,1]$, and by compactness take a finite set of $n$ such neighborhoods $U_i$ with corresponding functions $f_i$ which cover $[0,1]$.  We may assume that each function $f_i$ is positive on it's special neighborhood $U_i$ by multiplying by $-1$ if necessary.  Multiply each function $f_i$ by a continuous bump function $g_i$ which is positive on $U_i$ and zero outside of $U_i$.  Then we have functions $h_i=f_ig_i\in M$ which are positive on their special neighborhoods $U_i$ and zero outside.  Add these finitely many functions together to obtain a nowhere zero function $h=h_1+\dots+h_n\in M$.  

