We can then tile $\mathbb R$ with $|\mathbb R|$-many copies of $\mathbb Z$, coding different a different infinite sequence with our tile on each one. In fact, we can code sequences from different translation orbits on each copy of $\mathbb Z$, since there are $|\mathbb R|$-many of these orbits. This ensures that the tiling we end up with is aperiodic. Note that this gives an aperiodic tiling, but it does give an aperiodic monotile.

@TerryTao: If we allow reflections, I think it's possible to find an example of an aperiodic mono-tiling of $\mathbb R$. The idea is to use $\{1,5,6\}$ as a tile. Three copies of this tile can be used to cover $\{1,2,\dots,9\}$ in two distinct ways: there is a "two forwards one backwards" configuration and a "two backwards one forwards" configuration. Using these two configurations, we can tile $\mathbb Z$ in a way that codes an arbitrary infinite sequence of $0$'s and $1$'s.

An "aperiodic monotile" in the plane means a single tile that admits a tiling of the plane, but no periodic tiling. It seems you're asking for something much weaker in $\mathbb R$: a single tile that admits an aperiodic tiling (but it may also admit periodic tilings). In my opinion, this weaker version is still a good question, but the title is misleading.

@FrançoisG.Dorais: Here's an idea for a counterexample. Take two parallel lines in the plane, say $y=0$ and $y=1$, and connect them with a line segment, let's say $\{0\} \times [0,1]$. Let $d$ denote the taxicab metric on this set, and then let $X$ denote the subspace consisting of just the two lines (without the connecting segment).

@user42355: I'm just now seeing your comment. Yes, this works! I wish I would have thought of this. (Instead, I've found a different way around the issue in the answer I just posted.)

@JoelDavidHamkins: That's an interesting idea. But I didn't know about the Tarski invariants before today, and at a first glance I'm not sure how they interact with embeddings.

I like this question. I think it's likely to be tricky because often one shows a non-embedding of $A$ in $C$ by showing some cardinal invariant of the algebra to be smaller in $C$. But of course, we can't decrease a cardinal invariant infinitely often to solve this question.

@YCor: I read the problem differently from you. I think Dominic is asking whether the ordinal $\mathfrak{c}$ (defined as the least ordinal with cardinality $2^{\aleph_0}$) can be order-embedded . Dominic, perhaps you can clarify?

@DanTuretsky: I don't think so. Suppose $A \subseteq \mathbb R$ is Borel and $f: A \rightarrow [0,1]$ is a continuous surjection. If $G$ is (for example) a $G_\delta$ subset of $[0,1]$, then $f^{-1}(G)$ is a relatively $G_\delta$ subset of $A$, but that doesn't make it actually a $G_\delta$ set (in $\mathbb R$) -- just the intersection of a $G_\delta$ set with $A$.