Looking for rigor on details of a proof that a maximal ideal in $C[0,1]$ is of the form $M_\gamma=\{f \in R \mid f(\gamma)=0\}$ 
Let $R$ be the ring of real-valued continuous functions on the closed unit interval. If $M$ is a maximal ideal of $R$, prove that there exists a real number $\gamma \in [0,1]$ such that $M=M_\gamma=\{f \in R \mid f(\gamma)=0\}$.

Suppose that $M \subset R$ is a maximal ideal and that $M \ne M_\gamma$ for any $\gamma$. Then for all $\gamma \in [0,1]$ there exists a map $f_\gamma \in M$ such that $f_\gamma(\gamma) \ne 0$.
By continuity, if $f_\gamma$ does not vanish at $\gamma$, then there is an open ball $B_\gamma$ where $f_\gamma$ does not vanish. These open balls cover the compact set $[0, 1]$, hence some finite subset of them, say $B_{\gamma_1}, . . . , B_{\gamma_t}$ also covers [0, 1].
Now $f = f^2_{\gamma_1} + . . .+ f^2_{\gamma_t}$ is continuous and nowhere zero, so it is is unit in $R$. Since $f \in M$, we have that $M = R$.

Few remarks on this proof. First the part

By continuity, if $f_\gamma$ does not vanish at $\gamma$, then there is an open ball $B_\gamma$ where $f_\gamma$ does not vanish.

seems a bit handwavy and I would like to know why is this true?

Second part is the usage of compactness, but the way the pulled $f = f^2_{\gamma_1} + . . . + f^2_{\gamma_t}$ out of thin air is not natural to me. Is there a natural way to think about this map? And why do we even need the finite subcover here?

Lastly how is a continuous nowhere zero function automatically a unit in $R$?
 A: If $f$ is continuous at a point $\gamma$ and nonzero at that point, pick $\epsilon=|f(\gamma)|\gt 0$. Then there is such $\delta\gt 0$ such that, whenever $|x-\gamma|\lt\delta$ it is true that $|f(x)-f(\gamma)|\lt\epsilon=|f(\gamma)|$. Use the triangle inequality to conclude
$$|f(x)|=|f(\gamma)-(f(\gamma)-f(x))|\ge |f(\gamma)|-|f(\gamma)-f(x)|\gt 0$$
so $f(x)$ is also nonzero.

As for the map $f=f_{\gamma_1}^2+\ldots+f_{\gamma_n}^2$ i.e. $f(x)=f_{\gamma_1}^2(x)+\ldots+f_{\gamma_n}^2(x)$: the important facts about it:

*

*It is obviously nonzero, as sum of squares of some real numbers can only be zero when all of them are zero. On the other hand, at every point on the interval $[0,1]$ at least one of those functions $f_{\gamma_i}$ will be nonzero.

*Thus, it is a unit (invertible). Its (multiplicative) inverse is just $g(x)=\frac{1}{f(x)}$. A reciprocal of a continuous function is continuous wherever defined, i.e. wherever the denominator is nonzero, which in this case is for all $x\in[0,1]$.

*$f$ is a finite linear combination of elements in $M$ (that is why finiteness is important!), so is itself in $M$. But, as known from algebra, the only ideal containing a multiplicative unit in a ring is the whole ring.

