Ideal in a ring of continuous functions

One way to approach this is to show directly that the intersection of all zero sets of functions in $I$ is nonempty.

Denote the zero set of $f$ as $z(f)$. Of course $z(f)$ is always closed and nonempty for $f\in I$ (If it's empty, the function is a unit.)

The next observation is that the intersection of two zero sets for functions $I$ is nonempty. If to the contrary $z(f)\cap z(g)=\emptyset$, then $z(f^2+g^2)=\emptyset$, but this isn't possible since $f^2+g^2\in I$. From this proof you can see that for every $f$ and $g$ in $I$, $f^2+g^2$ is another element of $I$ whose zero set is just $z(f)\cap z(g)$. By induction, the intersection of finitely many zero sets is again a nonempty zero set of a function in $I$.

Now I claim that in a compact space, $\cap\{z(f)\mid f\in I\}$ is nonempty. If it were empty, then the complements of the $z(f)$ form an open covering of the space. Extracting a finite subcover, we have $\cap_{i=1}^n z(f_i)=\emptyset$. But we've already said this isn't possible, by the observations in the last paragraph.

Therefore $\cap\{z(f)\mid f\in I\}\neq\emptyset$, and any $\lambda$ in this set will do the trick for your question.

Since $[0,1]$ is compact, all maximal ideals look like $M_\lambda$ (see, for example, these notes). Now your question follows from the fact that any ideal is contained in a maximal ideal.

If possible, let $I\not\subset M_\lambda$ for all $\lambda$. Then for each $\lambda$ there will be a function $f_\lambda\in I$ such that $f_\lambda(\lambda)\neq 0$. By continuity there is a neighborhood $V_\lambda$ of $\lambda$ where $f_\lambda$ is nonzero. These neighbourhoods cover your compact set $[0,1]$. So there are $V_1,..,V_n$ covering $[0,1]$ and the function $g=f_1^2+...+f_n^2$ (tired of typing 'lambda' over and over again) therefore does not vanish anywhere in the interval. And it is an element of $I$. Since it doesn't vanish, $\frac{1}{g}$ is an element of your ring. It follows that $1\in I$ which is a contradiction.