<s>[INST] Can anything interesting be said about this fake proof? The Facebook account called BestTheorems has posted the following. Can anything of interest be said about it that a casual reader might miss?

Note that \begin{align}
\small 2 & = \frac 2{3-2} = \cfrac 2 {3-\cfrac2 {3-2}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}} \\[10pt]
\text{and } \\
\small 1 & = \frac 2 {3-1} = \cfrac 2 {3 - \cfrac 2 {3-1}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}} \\
\text{So } & 2=1.
\end{align}

 [/INST] $$x = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}}$$
$$x = \frac 2 {3 - x}\\
x^2 - 3x + 2 = 0\\
(x-1)(x-2) = 0$$
$1,2$ are both solutions.  However, if we consider this recurrence relation:
$$x_n = \frac 2 {3 - x_{n-1}}$$
when $x_{n-1} <1 \implies x_{n-1}<x_n<1$
And the squence converges to 1.
And
$$1<x_{n-1} < 2 \implies 1<x_n  <x_{n-1}$$
and the sequence again converges.
but,
$$2<x_{n-1} < 3 \implies x_{n-1}<x_n$$
and
$$x_{n-1} >3 \implies x_n< 0$$
The sequence isn't stable in a neighborhood of $2.$
and it converges to $1$ for nearly all starting conditions.
</s>