I only have to prove that the ideal $I$ of the ring of continuous functions from $\mathbb{R} \rightarrow \mathbb{R}$ is maximal where $I = \{f \in R \mid f(0) = 0\}$ (I know that his is true for any ideal where $\{f(x) \mid f(x_0) = 0\}$ for some $x_0$. Rather than using that $R/ I$ is a field iff $I$ is maximal, I wanted to try to practice proving it from the definition, which is that there is no ideal $J$ which $I$ is a proper subset of.
So take an element in $J$, $p(x)$, and since $I \subseteq J$, $p(0) = 0$ and so $p(x) \in I$, so $I = J$. I feel as if I have missed something crucial here since it seems quite short and non rigorous.
It's not non-rigorous per se, but it does have a non sequitur.
$I\subseteq J$ does not imply that $p(0)=0$. (In fact, it is the other way around: if $J\subseteq I$, then $p(0)=0$.) If $p\in J\setminus I$, then $p(0)\neq 0$.
You will have to proceed a different way. The burden is upon you to take an element $p\notin I$ and then explain why $(I,p)=C(\mathbb R,\mathbb R)$, or equivalently that you can generate a unit from elements of $I$ and multiples of $p$.
Hint: Let $f\in I$ be a function that's zero precisely at $0$. Then $f^2\in I$ and is positive except at $0$. For $p\notin I$, $p^2$ is nonnegative and $p^2(0)\neq 0$.