Verification of a proof that the ring of continuous functions on [0,1] is not noetherian

Verification of a proof that the ring of continuous functions on [0,1] is
not noetherian

My problem is to show that the ring of continuous real-valued functions on
the interval $[0, 1]$ is non-noetherian. I'd like to know if this argument
holds:
Let $I$ be the set of functions such that for every $f \in I$, there
exists an open neighborhood $[0, t)$ of $0$ such that $f(x) = 0$ whenever
$x \lt t$. Let's call the largest such neighborhood the zero neighborhood
of $f$.
This set is an abelian group, since the zero function obviously has such a
neighborhood, and given any pair of function $f$ and $g$, we can intersect
their zero neighborhoods to get the zero neighborhood of the sum. Lastly,
this set is closed under multiplication in $\mathbb{R}$: scaling by $0$
gives you the zero function, and scaling by anything else preserves the
zero neighborhood.
Then, to prove our ring non-noetherian, we must show this ideal cannot be
finitely generated. Suppose it was, $I = (f_1 f_2 ... f_k)$ with each
$f_i$ having a zero neighborhood of $[0, x_i)$. We will find a function
$f$ which cannot be written as a finite linear combination of these $f_i$.
We noted above that closure under addition and under scaling may only
affect the zero neighborhood in certain ways. Addition of two nonzero
functions with zero neighborhoods $[0, x_1)$ and $[0, x_2)$ will leave you
with a function whose zero neighborhood is $[0, \min_{ x_1, x_2} )$.
Scaling by a nonzero real will preserve the zero neighborhood. This means
that the linear combination of $f_i$ must necessarily either have $[0, 1]$
as its zero neighborhood, or else it must have $[0, x_i)$ for some $i$.
To construct a counterexample, we define $\widetilde{x}$ to be the
midpoint between $\max {x_i}$ and $1$. Then we let $f(x)$ be uniformly
zero until $x_i$, then let it grow linearly from there on out.
Wordy, but hopefully clear. There are some subtle points I'm missing. For
one, I know that I would need to exclude the zero function from the final
$\max {x_i}$ calculation. I also need to show that if a continuous
function is zero on $[0, 1)$, it is the zero function. These two guarantee
the choice at the end is legitimate.
Does this seem like a legitimate approach? Or did I miss anything?