If $(X,T_d)$ is a complete metric space and we have a sequence of closed subsets ${K_n}$ s.t. $\operatorname{int}(K_n)$ is empty, then
$$\operatorname{int}\Big(\bigcup_{n=1}^{\infty}K_n\Big)=\emptyset.$$
I searched about Baire's theorem but it said something different, with open sets and the intersection, is it the negation of Baire's theorem?
Thanks in advance
The Baire category theorem states the following:
Now set $U_n = K_n^c$. Then $U_n$ is a sequence of open sets and $\overline{U_n} = \overline{K_n^c} = ((K_n)^\circ)^c = \emptyset ^c = X$. Now from the Baire category theorem we know $$X = \overline{\bigcap U_n} = \overline{\big(\bigcup K_n\big)^c} = \Big(\big(\bigcup K_n\big)^\circ\Big)^c,$$ i.e. $\big(\bigcup K_n\big)^\circ = \emptyset$.