Are compact subsets of $\mathbb{R}$ with the compact complement topology closed?

Given the usual topology $\tau$ on $\Bbb{R}$, define the compact complement topology on $\mathbb{R}$ to be $$\tau'=\{A\subseteq \Bbb{R}:A^C\text{ is compact in }(\Bbb{R},\tau)\} \cup \{\emptyset \}.$$ Are compacts sets in $(\Bbb{R},\tau')$ closed (in $(\mathbb{R} , \tau')$)?

After one hour's struggle I could not prove this. I can see that all compact sets in $(\Bbb{R},\tau)$ are still compact in $(\Bbb{R},\tau')$ and they are trivially closed, but this is certainly not equivalent to the claim to prove.

You are correct. Clearly $\Bbb Z$ is not $\tau'$-closed, since it’s not $\tau$-compact, but I claim that it is $\tau'$-compact. Let $\mathscr{U}$ be a $\tau'$-open cover of $\Bbb Z$. Fix $U_0\in\mathscr{U}$; then $\Bbb R\setminus U_0$ is $\tau$-compact and hence bounded. But then $\Bbb Z\setminus U_0$ is bounded and therefore finite, so only finitely many members of $\mathscr{U}$ are needed to cover it.