Some years ago, I came up with this false proof of the irrationality of $\pi$.

It suffices to prove that $x=\pi-3$ is irrational.

For real $y$ with $0\le y\lt1$,
and positive integer $j$, define $d_j(y)$ to be the $j$th digit in the
decimal expansion of $y$.

Let $r_1,r_2,\dots$ be
an enumeration of the rationals in $[0,1)$. The $\it value$ of this
enumeration is $n$ if $d_n(r_n)=d_n(x)$ and $d_j(r_j)\ne d_j(x)$ for $j\lt n$.
If there is no such $n$, then the value of the enumeration is infinite. Note
that if there is an enumeration of infinite value, then $x$ is irrational; it
cannot equal any of the enumerated rationals, as it differs from the first
rational in (at least) the first decimal place, from the second in the second,
etc.

Note also that there are enumerations of arbitrarily large value. For, given
any $n$, you can find $n$ rationals such that the first differs from $x$ in
the first decimal, the second differs from $x$ in the second decimal, and so
on, and then any enumeration that starts off with these $n$ rationals will
have value greater than $n$.

Now, the set of all enumerations of the rationals can be partially ordered by
value; if $E_1$ and $E_2$ are enumerations, then $E_1>E_2$ if the value of
$E_1$ exceeds the value of $E_2$. By Zorn's Lemma, there is an enumeration
maximal with respect to this order. This maximal enumeration cannot have a finite
value --- as we have seen, there are enumerations of arbitrarily great finite
value. So, it must have infinite value. So, $x$ is irrational.

An alternative use for this argument is to apply it to prove that $1/3$ is irrational, the contradiction with the known rationality of $1/3$ thereby establishing that Zorn's Lemma is false.