Can every (non-odd) set be partitioned into subsets of two elements? By this I mean that given a set $A$, there is some $S \subset \mathscr{P}(A)$ such that every element of $S$ has exactly two elements and every element of $A$ appears in exactly one element of $S$. For a trivial example, $\mathbb{Z}^+$ can be partitioned into $\{ \{1,2\}, \{3, 4\}, \dots \}$.
It seems like "common sense" that this partition should exist, but I have been unable to prove it. I was hoping for a construction using simple set theory operations, but a proof by negation would be second-best. However I would be happy just to know if this is something that can be proved only by some more advanced mathematics.
My thoughts on constructing a partition for progressively more complicated sets $A$ follow.
If $A$ were allowed to be finite, it's easy to see that it can be so partitioned depending on whether the cardinality is even (can be partitioned) or odd (cannot be partitioned).
If $A$ were countably infinite, then one could use recursion to build up the required partition. I guess it would look something like this: If $f: \mathbb{Z}^+ \rightarrow A$ is a counting function, then let $x = f(1)$, choose any $y = A - \{x\}$, and set $S_1 = \{\{x,y\}\}$. Then for subsequent iterations, let $x = f(\min \{i \vert f(i) \in A - S_n\})$ and $y$ be any other element of $A - S_n$ and then define $S_{n+1} = S_n \cup \{ x, y \}$.
Now if $A$ is uncountable but has an order (which I assume means that there is a bijection with $\mathbb{R}$, although I don't know this for a fact), then a partition might be the image under this bijection of $$\bigcup_{i\text{ even}} \{\{x, x+1\} \vert x \in [i, i+1)\}$$
But now we come to the general case of uncountably infinite sets. I'm not a mathematician and so I have pretty limited tools at my disposal, but my failure to produce a proof so far has led me to start questioning my "common sense" assumption that said partition exists.
The answer to your question is yes, but my proof will rely heavily on the Axiom of Choice. I don't know immediately whether your claim is true without choice.
Well order $A$ in the order type of its cardinality. That means (using consequences of the Axiom of Choice) that we pick a bijection $f: \kappa \to A$ for some cardinal $\kappa$. Let $\alpha \in \kappa$ be any ordinal. Then $\alpha$ can be uniquely expressed as $\beta + n$, where $\beta$ is a limit ordinal and $n \in \Bbb N$. Pair up $f(\beta+ 2k)$ with $f(\beta+2k+1)$.
When $A$ is uncountable, this proof essentially partitions $A$ into a whole bunch of copies of $\Bbb N$, and then uses the obvious partition on $\Bbb N$.