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Where $a<b$, say that the four “types” of non-empty bounded intervals are:

$(a,b)$, $[a,b]$, $(a,b]$, and $[a,b)$.

Let $\langle X,< \rangle$ and $\langle Y,< \rangle$ be dense linear orderings without endpoints such that

  1. $\langle \mathbb{R},< \rangle$ is order-embeddable into $\langle X,< \rangle$ and into $\langle Y,< \rangle$,

  2. all the non-empty bounded intervals of $\langle X,< \rangle$ of the same type are order-isomorphic, and

  3. all the non-empty bounded intervals of $\langle Y,< \rangle$ of the same type are order-isomorphic.

What are some examples of $\langle X,< \rangle$ and $\langle Y,< \rangle$ such that $ZFC$ proves that they are incomparable (neither is order-embeddable into the other)?

What are some examples of $\langle X,< \rangle$ and $\langle Y,< \rangle$ for which it is independent of $ZFC$ whether they are incomparable?

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As Wojowu says, the long ray and the reverse long ray are each locally like the reals, but one has an increasing $\omega_1$-sequence and no decreasing $\omega_1$-sequence, and the other a decreasing and no increasing $\omega_1$-sequence, and this prevents an embedding in either direction. So they are incomparable.

One can turn this example into an answer also to question 2. Namely, let $X$ be the long ray, that is, $X=[0,1)\times\omega_1$, and let $Y=([0,1)\times\omega_1^L)^*$, which is like the inverse long ray, but the length of it is $\omega_1$ as determined by the constructible universe $L$. Each of these orders is locally like the reals, but the comparability of these orders is independent of ZFC. If $V=L$, then the orders are just as in the previous example, and so they are incomparable. But if $\omega_1^L$ is countable, then $Y$ is isomorphic to an interval in the reals, which embeds into $X$. So their comparability is independent of ZFC.

In the comments, the OP has requested examples which furthermore are symmetric, in the sense that they are isomorphic to their inverse. If they are also to be locally like $\mathbb{R}$, then there are no examples.

Theorem. If a linear order has all bounded intervals isomorphic to an interval of the reals, then it is isomorphic to one of the following:

  1. An interval of the reals.
  2. The long ray. (including or excluding the least point)
  3. The inverse long ray. (including or excluding the largest point)
  4. The long line.

Proof. Since every bounded interval of the order is isomorphic to an interval in the reals, there are no bounded increasing or decreasing $\omega_1$-sequences. From this, it follows that the cofinality of the order is at most $\omega_1$, and the co-initiality as well. So the four possibilities arise from the four possible combinations of cofinalities at the top and the bottom. QED

Corollary. If $X$ and $Y$ are linear orders that have all bounded intervals isomorphic to intervals of the reals, and if at least one of them is symmetric, then they are comparable by embeddability.

Proof. If one of the orders is empty, then it embeds into the other. If either is a singleton, then it embeds. So we may assume they each have at least two points, and so they each have nontrivial real intervals. If $X$ is symmetric, then either it is isomorphic to an interval in the reals, in which case it embeds into $Y$, or it is the long line, in which case $Y$ embeds into it. QED

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    $\begingroup$ Perhaps you want to say that $Y=[(\omega_1)^L \times [0,1)]^*$. Otherwise $Y$ is not homeomorphic to an interval of reals. $\endgroup$
    – Ramiro de la Vega
    Commented Sep 5, 2016 at 15:35
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    $\begingroup$ This answer reminds me of this one: mathoverflow.net/q/77312. One must accept it but it is probably not exactly what the OP was looking for :) $\endgroup$
    – Ramiro de la Vega
    Commented Sep 5, 2016 at 15:37
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    $\begingroup$ @RamirodelaVega One can make similar examples to the one in my answer by using any cardinal $\kappa$ that is provably $\omega$ or $\omega_1$, but independent of ZFC which one. There are abundant examples of that, and you might find some of them more attractive. $\endgroup$
    – Joel David Hamkins
    Commented Sep 5, 2016 at 15:50
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    $\begingroup$ @JoelDavidHamkins, thank you! Are there examples in which the first order does not differ with respect to the length of increasing vs. decreasing sequences it contains, and similarly for the second order? $\endgroup$
    – Haidar
    Commented Sep 5, 2016 at 15:50
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    $\begingroup$ Ah, perhaps you want them each to be isomorphic to their inverse order? $\endgroup$
    – Joel David Hamkins
    Commented Sep 5, 2016 at 16:03

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