Proving that an uncountable subset of $\mathbb{R}$ has an strictly increasing sequence. As the title says, I need to prove that given $A\subseteq\mathbb{R}$ an uncountable subset ($|A|>\aleph_0$) then there exist $\{a_n\mid n\in\mathbb{N} \}\subseteq A$ such that $a_1<a_2<a_3<\cdots$, .i.e., $A$ has an strictly increasing sequence.
My attempt:
We claim that there exist $x\in A$ such that $(x,\infty)\cap A$ is an uncountable set. Suppose not, i.e., for all $x\in A$ we have that $(x,\infty)\cap A$ is countable. Then, we have two cases:

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*$A$ has a lower bound. Then there exist $\inf(A)$. We can take a decreasing sequence $\{ b_n\}\subseteq A$ such that $b_n$ converges to $\inf(A)$. Then $A=\displaystyle\bigcup_{n\in\mathbb{N}}[(b_n,\infty)\cap A]$. But $(b_n,\infty)\cap A$ is countable. Therefore $A$ is countable. This is a contradiction.

An important detail is that we have the case where $\inf(A)\in A$ and the case where $\inf(A)\notin A$. If $\inf(A)\notin A$ the above argument works. If $\inf(A)\in A$ then $A=\left(\displaystyle\bigcup_{n\in\mathbb{N}}[(b_n,\infty)\cap A] \right)\cup \{ \inf(A)\}$ and, again, $A$ is countable.


*$A$ has no lower bound. Then we can take an strictly decreasing sequence $\{b_n \}\subseteq A$ such that $\lim\limits_{n\to \infty}b_n=-\infty$. Then $A= \displaystyle\bigcup_{n\in\mathbb{N}}[(b_n,\infty)\cap A]$ and, again, $A$ is countable.

Therefore, there exist $x_1\in A$ such that $(x_1,\infty)\cap A$ is uncountable. The set $(x_1,\infty)\cap A$ being uncountable, satisfies the hypothesis and therefore, there exist $x_2\in(x_1,\infty)\cap A$ such that $(x_2,\infty)\cap(x_1,\infty)\cap A=(x_2,\infty)\cap A$ is uncountable. Now we have $x_1,x_2\in A$ such that $x_1<x_2$. Following this construction we obtain the desired result.
Is my proof correct? Any suggestion?
 A: Yes, this is correct. It can be simplified however: just note that $$A=\bigcup_{z\in\mathbb{Z}}A\cap [z,\infty),$$ and so for some $z\in\mathbb{Z}$ we must have $A\cap [z,\infty)$ be uncountable. This gets rid of the case breakdown, and also shows that we don't need to talk about infs/sups (which means we can apply this to linear orders which are more complicated than $\mathbb{R}$, to a certain extent).

Actually, there is one slight issue: the result you state

If $A\subseteq\mathbb{R}$ is uncountable then there is some $x$ such that $A\cap (x,\infty)$ is uncountable

isn't really phrased in such a way that it can be "iterated" the way we want. Rather, what we really should say is:

If $A\subseteq (a,\infty)$ is uncountable then there is some $b\in (a,\infty)$ such that $A\cap (b,\infty)$ is uncountable.

The proof of this is basically the same as the above, though: just let $f:\mathbb{R}\rightarrow (b,\infty)$ be your favorite order-preserving bijection, and think about $\{f(z):z\in\mathbb{N}\}$.
A: The proof is correct, but you’ve done some unnecessary work in the first part, and the last part could be tightened up quite a bit by making the recursive construction of the sequence a bit more rigorous.
To prove that there is an $x\in\Bbb R$ such that $A\cap(x,\to)$ is uncountable, just observe that if not, then $A=\bigcup_{n\in\Bbb Z}\big(A\cap(n,\to)\big)$ is the union of countably many countable sets and is therefore countable.
For the second part, let $x_0\in\Bbb R$ be such that $A\cap(x_0,\to)$ is uncountable. Given $x_n\in\Bbb R$ such that $A\cap(x_n,\to)$ is uncountable, the first part ensures that there is an $x_{n+1}\in A\cap(x_n,\to)$ such that $A\cap(x_{n+1},\to)$ is uncountable. Clearly $x_n<x_{n+1}$ for each $n\in\Bbb N$, so $\langle x_n:n\in\Bbb N\rangle$ is a strictly increasing sequence in $A$.
