Question on Baire lemma : why a non empty complete metric space have a non-empty interior?

Baire's lemma says :

Let $X$ a complete metric space. Let $(X_n)_n$ a sequence of closed subset. Suppose $Int(X_n)=\emptyset$ for all $n$. Then $Int(\bigcup_{n\in\mathbb N}X_n)=\emptyset$.

And here an equivalent form :

Let $X$ a non-empty complete metric space. Let $(X_n)_n$ a sequence of closed set s.t. $\bigcup_{n=1}^\infty X_n=X$. Then there is a $n_0$ s.t. $Int(X_n)\neq \emptyset$.

For example, $\{1\}$ with the induced topology from $\mathbb R$ is a complete metric space, we can take $X_n=\{1\}$ for all $n$, we have that $\bigcup_{n=1}^\infty X_n=X$, but $Int(X_n)=\emptyset$ for all $n$. Where is my mistakes here ?

But $Int(X_n)=X_n$, because for $1$ there is an open neighbourhood namely $X_n$ with $X_n\subseteq X_n$.