Where is the mistake in the argument in favor of the (erroneous) claim "every Dedekind cut is a rational cut"?

A cut is a set $C$ such that:

(a) $C\subseteq \mathbb Q $

(b) $C \neq \emptyset $

(c) $C \neq Q $

(d) for all $a, c \in \mathbb Q $ , if $c\in C$ and $a\lt c$ , then $a\in C $

(e) for all $c\in \mathbb Q$ , if $c\in C$, then there exists $d\gt c$ such that $d\in C$

By a rational cut is meant a cut in $\mathbb Q$ of the form $C(r) = \{c | c\lt r, r\in \mathbb Q\}$.

I know that the statement "every cut is a rational cut" can be disproved by giving a counterexample. However, I can't prevent myself from finding an air of plausibility to the following argument aiming at justifying that every cut is of the form $C(r)$.

Hence my request: can you please show me where the following argumentgoes wrong?

Step 1 :

Suppose $C$ is a cut , and that $C$ is not of the form $C(r)$ form some $r \in \mathbb Q$.

This amounts to saying:

(1) there is no $r$ such that for all $c$ if $c \in C$ then $c\lt r $

(2) implying that: for all $r$ it is not the case that for all $c$ if $c \in C$ then $c\lt r $

(3) implying that: for all $ r$ there is some $c$ such that $c\in C$ and $c\geq r $

Step 2:

Suppose $r\in \mathbb Q$ and $r\notin C $.

By (3) above there is some $c$ such that: $c\in C$ and $c\geq r$.

Now, if $c=r$ then $r\in C$.

Also, if $c\gt r$ then $ r\lt c$ and, by the "cut" definition, $r\in C$.

So the hypothesis $r \notin C$ leads to a contradiction; meaning that $\mathbb Q\subseteq C$, since $r$ was arbitrary.

Finally we have $C\subset \mathbb Q$ and $Q \subseteq C$, a contradiction (since the first statement negates that all rationals belong to $C$ while the second asserts the same proposition).

The conclusion seems to be that a Dedekind cut that is not a rational cut is not a Dedekind cut; in other words, every Dedekind cut has to be a rational cut.

Note: I can see an objection to the conclusion of "step 2", namely that what what the argument shows is that the first conjuction is false , namely $r\in \mathbb Q$. But how to make sense of the negation of this conjunct since the set of rationals is the domain over which range all our number variables (in this argument)?