In this answer, @Hyperplane defined an infinite set as follows:
Well a set $M$ is infinite if for every finite subset $U \subsetneq M$ there exists a $x \in M, x\notin U$. Consequently, you will be able to construct a sequence $(x_n)_{n \in \mathbb{N}} $ of distinct elements in $M$. Therefore $|M| \ge \aleph_0 $
Then @Andrés E. Caicedo commented here:
This is not true as written. Replace $\subsetneq$ with $\subseteq$, and it will be fine now.
I think that @Hyperplane's definition of an infinite set is wrong. My reasoning is:
For every subset $U \subsetneq M$, we always have $\exists x \in M, x \notin U$ by the definition of $\subsetneq$. So @Hyperplane just restated the property of $\subsetneq$ and said nothing about the definition of infinite set.
When I replace $\subsetneq$ with $\subseteq$ in @Hyperplane's answer as suggested by @Andrés E. Caicedo, I even find the new definition more awkward. My reasoning is:
For $U=M$, there does NOT exist any $x \in M, x\notin U$. Thus the statement Well a set $M$ is infinite if for every finite subset $U \subseteq M$ there exists a $x \in M, x\notin U$ is wrong. Let alone define an infinite set.
Could you please check whether my above reasonings are correct?
Your reasoning about the original failed definition is correct.
Your critique of the modified definition misses the point slightly. You seem to be missing the condition that $U$ has to be finite; so if $M$ is not finite you cannot choose $U=M$ as a finite subset of $M$.
On the other hand, the outcome of this is that the modified definition simply says that $M$ is infinite iff $M$ is not a finite subset of $M$. Or in other words:
If we already know what "finite" means, this is a good definition. But bringing non-proper subsets into it just muddies the waters.
(On the other hand, if we don't already have a definition of "finite", then neither of the proposed definitions make sense).