Uncountable closed set of irrational numbers

If enumerate the rationals $\mathbb{Q}=\{q_1,q_2,\ldots\}$ and define the intervals $I_n=(q_n-2^{-n},q_n+2^{-n})$, we have that the set $S=\bigcup_{n=1}^\infty I_n$ contains every rational number and $$m(S)\leq\sum_{n=1}^\infty m(I_n)=\sum_{n=1}^\infty 2^{-\left(n-1\right)}=2$$ Therefore, the complement of $S$ in $\mathbb{R}$ is of infinite measure (hence uncountable) and contains only irrationals.

List the rationals as $r_0,r_1,\dots,r_n,\dots$.

Around the $n$-th rational, make an open interval of width $\epsilon/2^n$. Take the union of these. The complement does the job.

The irrational numbers are homeomorphic to $\Bbb N^{\Bbb N}$. Let $C=\{0,1\}^{\Bbb N}$, viewed as a subset of $\Bbb N^{\Bbb N}$. By the Tikhonov product theorem $C$ is compact, and $\Bbb N^{\Bbb N}$ is Hausdorff, so $C$ is closed in $\Bbb N^{\Bbb N}$. (In fact $C$ is homeomorphic to the middle-thirds Cantor set.) Finally, $C$ is clearly uncountable. Now if $h:\Bbb R\setminus\Bbb Q\to\Bbb N^{\Bbb N}$ is any homeomorphism, $h^{-1}[C]$ is an uncountable closed subset of $\Bbb R\setminus\Bbb Q$, the irrationals. Being compact, it’s also closed in $\Bbb R$.