I found out that $\aleph_1$ is an upper bound. I can show that any well ordered subset of $ \mathbb R$ is countable, so its order type is an element of $\omega_1$. I'd like to show that $|X|= \aleph_1$, but I don't know how to do it. Some hint?

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    $\begingroup$ In fact, every conutable ordinal can be embedded into $\mathbb{Q}$. We can prove that every countable linearly oredered set is embedded into $\mathbb{Q}$ by back-and-forth. $\endgroup$ – Hanul Jeon Jun 18 '17 at 14:13
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    $\begingroup$ @HanulJeon Nitpick: You need "back-and-forth" to prove that every countable dense linear order with no first or last element is isomorphic to $\mathbb Q.$ To prove that every countable linear order is embeddable in $\mathbb Q$ you don't need "back-and-forth", just "forth". $\endgroup$ – bof Jun 23 '17 at 4:09
  • $\begingroup$ @bof. We can also avoid model theory altogether and use elementary means (and more time!). The result that a non-empty dense countable linear order without end-points is isomorphic to $\mathbb Q$ is due to Cantor in the 19th century. And it is easy to embed any other countable linear into such an order. $\endgroup$ – DanielWainfleet Jun 23 '17 at 9:14

Hint: Try proving that any countable ordinal $\alpha$ embeds in $\mathbb{R}$ by induction on $\alpha$. You'll make use of the countability assumption at limit stages, in that any countable limit ordinal has countable cofinality.

Alternatively, you can prove that any countable totally ordered set $X$ at all embeds in $\mathbb{R}$, by enumerating the elements of $X$ and defining an embedding $f:X\to\mathbb{R}$ on one element at a time.

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    $\begingroup$ Alternatively, you can prove that any countable totally ordered set $X$ embeds in $\mathbb R$ by choosing an injection $\varphi:X\to\mathbb N$ and defining an embedding $f:X\to\mathbb R$ by writing $$f(x)=\sum_{y\lt x}2^{-\varphi(y)}.$$ $\endgroup$ – bof Jun 18 '17 at 1:50
  • $\begingroup$ @bof Do you know of a way to modify your embedding to ensure the range is contained in $\mathbb Q$? $\endgroup$ – Andrés E. Caicedo Sep 3 '17 at 5:10
  • $\begingroup$ @AndrésE.Caicedo Nope. $\endgroup$ – bof Sep 3 '17 at 5:28

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