Noetherian hypothesis when permuting elements of a regular sequence? One of the first results showed when studying regular sequences is that we are allowed to shuffle the elements of the sequence if the ring is noetherian, local, and the module is finite (see Proposition 2 in here for a proof). 
I know the ring being local and the module being finite are required to use Nakayama's lemma, but I can't spot where is noetherianity used!
Edit: By MooS' comment I realize Akhil and probably all the other sources are using Krull's Intersection theorem, which I thought I didn't need for 'my' proof (I learned it from Bruns-Herzog). So now I wonder what is wrong with the following:
We want to show  that under all the hypothesis, if $x,y$ is an $M$-sequence, then $y,x$ is an $M$-sequence. Let's show that $y$ is not a zero-divisor.
Denote the kernel of multiplication by $y$ on $M$ by $K$. Assume $m\in K$. Then, by the regularity of the sequence $m\in xK$ (use $\overline{ym}=0\in xM$) and we can write $m=xm'$ for some $m'\in K$, i.e., $K\subseteq xK$. Conversely, if $xym'=0$ by regularity $ym'=0$ and $m'\in K$. Therefore $K=xK$, and by Nakayama's lemma (version from AM) $K=0$. Thus, $y$ is not a zero-divisor. QED
What is wrong with this proof?

The assumption that the ring be noetherian is used when Krull's intersection theorem is applied. And the assumption is necessary, as Stacks Project's tag 00LH shows, for example: consider $k[x,y,w_1,w_2,\ldots]/(yw_1,yw_2,\ldots,w_1-xw_2,w_2-xw_3,\ldots)$ and localise in the maximal ideal generated by $x,y$ and all the $w_i$, $i\in\mathbb{N}$. Then $x,y$ is a regular sequence, but $y$ is a zero divisor.
We can detect the issue with the OP's proof in the example: $K=\mathrm{Ann}(y) = (w_1,w_2,\ldots)$ does satisfy $xK=K$, but we can't conclude $K=0$ from Nakayama's lemma since  $K$ is not finitely generated. So it seems the proof is okay if the ring is noetherian and it does not use Krull's intersection theorem.
