There exists a partial ordering of $ \mathbb{R} $ with no uncountable chains or antichains here's an exercise I'm stuck upon.  I'm not sure how to approach this:

There exists a partial ordering of $ \mathbb{R} $ with no uncountable
  chains or antichains

While looking for some help, I came across the Suslin tree, which, so it seems, doesn't necessarily exist. I'm not sure where to look for some clues
I would appreciate some help
 A: If, by antichain, you mean all members are pairwise incompatible, then take $P$ to be the set of all finite partial maps from $\mathbb{R}$ to $\{0, 1\}$ ordered by reverse inclusion. This also gives examples of all cardinalities.
If by antichain, you mean all members are pairwise incomparable, then fix a well ordering $\preceq$ of $\mathbb{R}$ and for $a, b \in \mathbb{R}$, define $a \leq_1 b$ iff $a \leq b \wedge a \preceq b$. Note that, by Erdős-Rado, this example is optimal in the sense that there is no such partial ordering on a set of size $> |\mathbb{R}|$.
A: I am just going to post a more verbose version of the second paragraph of hot_queens answer (which is also basically this answer), since it took me some time to understand it:
Fix some well-ordering $\preceq$ of $\mathbb R$, and then define $a \sqsubseteq b$ iff $a \leq b$ and $a \preceq b$.
Then let $S$ be a chain, and suppose $S$ is uncountable. For every $y$, let $S_{<y}$ be the set of numbers in $S$ that are smaller (w.r.t. $\leq$) than $y$. Then, let
$$
x = \inf \{y : S_{<y}\text{ is uncountable}\}
$$
We have $x \neq \infty$ because $S$ is uncountable, and $x \neq -\infty$ because $S$ has a smallest element. So there are two cases:


*

*$x$ is the largest number such that $S_{<x}$ is countable. Then there has to an infinite descending sequence of numbers in $S$ that are greater than $x$. Because $S$ is well-ordered w.r.t. $\preceq$ and thus also w.r.t. $\sqsubseteq$, this is a contradiction.

*$x$ is the smallest number such that $S_{<x}$ is uncountable. Then there has to be an infinite ascending sequence of numbers $x_i \in S_{<x}$ that converge to $x$.
Now, we have $S_{<x} = \bigcup_{i \in \mathbb N} S_{<x_i}$. But every $S_{<x_i}$ is countable, so $S_{<x}$ has to be countable as well, which is a contradiction.
So every chain is countable.
The argument for antichains is extremely similar: Any antichain is well-ordered w.r.t. $\preceq$ and thus "reverse well-ordered" w.r.t. $\leq$. So you just repeat the previous argument while "flipping" every reference to the ordering $\leq$. 
