As suggested by @IzaakvanDongen this answer is a brief summary of what came out in the comments, therefore one can find further details on the results I am going to summarize in the comments above.
Question 1. A simple lower bound of $|S|$ can be made with the following injective map: $$\mathbb R \to S : r \mapsto \{]-\infty,r],]r,+\infty[\}.$$
For the upper bound, we notice that a partition of $\mathbb R$ in convex subsets can contain at most $\aleph_0$ (disjoint) convex subsets with cardinality $>1$, otherwise there'd be an injection from a set with cardinality $> \aleph_0$ to $\mathbb Q$, that is absurd. Now this means that every partition $P \in S$ can uniquely be mapped in an element of $\mathcal P_{\leq \aleph_0}(\{\text{convex subsets of $\mathbb R$}\})$ (and the latter has cardinality $2^{\aleph_0}$). Otherwise we can also have the upper bound with the injective map: $$S \to (((\mathbb R \cup \{\pm \infty\})^2 \times \{0,1\}^2) \cup \{\spadesuit\})^\omega : P \mapsto (a_i)_{i \in \omega} $$
where $(a_i)_{i \in \omega}$ encodes the convex subsets with cardinality $>1$ of $P$, and, when these subsets are in a finite number, the map
gives eventually $\spadesuit$ (this needs further details to be written properly, but I think the idea is pretty clear).
Question 2a. A first remark that's needs to be done is that a set is badly-ordered if and only if doesn't have a minimum element, so we're looking
for a countable total order that has a minimum element but at the same time has a non-empty subset without a minimum element.
Two simple examples can be done, the first taking the following subset of rationals:
$$\{-1\} \cup \left\{\frac 1n : n \in \omega \setminus \{0\}\right\}$$
the second one taking the sum of total orders:
$$(\{\pi\},<) + (\mathbb Q,<)$$
Both examples are countable total orders with a minimum element and a subset without minimum, so taking a bijection with $\mathbb N$, we can induce one of the order above on $\mathbb N$.
Question 2b. As un upper bound, the one given at the end of the question above is correct, now we provide a way to encode an countable binary sequence in a bad ordered set. We define:
$$\{0,1\}^\omega \to R : (a_i)_{i \in \omega} \mapsto A:= \mathbb Z \sqcup \bigsqcup_{i \in \omega} F(a_i)$$
where $F(a_i) = \mathbb Z$ if $a_i = 1$, otherwise $F(a_i) = \{i\}$. This is a countable sum of total orders, so it's a countable total order, and in particular, thanks to the first copy of $\mathbb Z$, it's a bad order. To prove that different sequences give us non isomorphic orders we can prove by induction that if we have $(a_i)_{i \in \omega} \mapsto A \cong B \leftarrow\!\shortmid (b_i)_{i \in \omega}$, then $\forall n \in \omega \ (a_i)|_n = (b_i)|_n$, which implies $(a_i)_{i \in \omega} = (b_i)_{i \in \omega}$. So we conclude that the map is well-defined and injective.
Best Answer
Well, consider $\Bbb R^2$ with the lexicographic ordering. This is not isomorphic to any subset of $\Bbb R$, since it can be partitioned into $2^{\aleph_0}$ nontrivial intervals.
However, if $(x_\alpha,y_\alpha)$ is a family of size $\omega_1$ in $\Bbb R^2$, then on at least one of the coordinates it has to violate monotonicity.
In general, the real numbers is the unique linear order which is:
So it means that a linear order can be mapped into $\Bbb R$ if and only if its completion is separable. Which holds, by the definition of completion, if the linear ordering is separable.
To see why this is indeed the case, pick a countable subset which is dense in the order, then that countable subset is mapped into the rationals, and by completeness of $\Bbb R$, we can now realize the rest of the linear ordering as well.
It follows from separability that every family of pairwise disjoint open intervals is countable. Call this property the countable chain condition.
Suslin asked in 1920 whether or not one can replace "separable" by "ccc" (although the terminology wasn't developed until much later). It turned out that the answer is independent of the usual axioms, i.e. ZFC. In other words, it is consistent that there is a linearly ordered space which is complete, dense, and ccc but not separable, and it is consistent that there is none.
And since we mention Suslin lines, let us also mention Aronszajn lines. Which are linear orders of size $\aleph_1$ but contain no uncountable subset which can be embedded in the real numbers, and no uncountable monotone sequence. The existence of these, unlike Suslin lines, is provable from ZFC.