Prove that any real number can be expressed as the sum of two irrational numbers
$\forall x\in\Bbb R$, the law of excluded middle tells us that $x$ must either be irrational or rational. This should be obvious since, after all, $x\in\Bbb Q$ or $x\in\Bbb R \setminus\Bbb Q$ by definition of rational and irrational.
If $x$ is irrational then $x = \frac{x}{2} + \frac{x}{2}$, where $\frac{x}{2}$ is irrational. This is clear since a rational, $\frac12$, times an irrational, $x$, is an irrational.
If $x$ is rational then $x = (\sqrt{2}) + (x-\sqrt{2})$, where both $\sqrt{2}$ and $x-\sqrt{2}$ are irrational. This is clear since a rational, $x$, added to an irrational $-\sqrt{2}$, is still irrational.
So all $x\in\Bbb R$ can be expressed as a sum of irrationals. $\;\;\;\blacksquare$
Try the general equality $x = (\frac12 x + z) + (\frac12 x -z)$, $\forall x \in\Bbb R$, where we choose an arbitrary $z\in\Bbb R$ such that $z\in\Bbb Q$ when $x\in\Bbb R\setminus \Bbb Q$ and $z\in\Bbb R\setminus\Bbb Q$ when $x\in\Bbb Q$. In this way both $(\frac12 x + z)$ and $(\frac12 x - z)$ are guaranteed irrational, and the sum is always $x$, regardless of which subset it comes from. If we further ensure that $z\ne 0$ then the two number forms are guaranteed distinct.
If $r$ is irrational, then $r=\frac{r}2+\frac{r}2$.
To tell the truth, for $r$ rational we also have $r=\frac{r}2+\frac{r}2$, but it is not useful in this case. :-)
A cardinality-based proof:
Let $r - \mathbb{Q} = \{r-q | q \in \mathbb{Q}\}$. This clearly has the same cardinality as $\mathbb{Q}$ hence is countable. Let $A = \mathbb{Q}' \, \backslash \,(r-\mathbb{Q})$ be the set of irrational numbers with $r - \mathbb{Q}$ removed. $A$ is clearly uncountable. For any $x \in A$, both $x$ and $r-x$ are irrational and sum to $r$. Thus, not only can $r$ be expressed as the sum of two irrational numbers, it can be so expressed in uncountably many ways. Since there are only countably many pairs $(x,y)$ which contain at lease 1 rational number and sum to $r$, it follows that most pairs of numbers which sum to $r$ have both terms irrational.
The same proof works if the set of irrationals is replaced by any cocountable set of reals (a set whose complement is countable). For example, given any real number $r$, most pairs of numbers which sum to $r$ consists of two non-computable numbers.