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The theorem statement is as follows.

Let $X$ be a complete metric space and let $(X_n)_{n \geq 1}$ be a sequence of closed subsets in $X$. Assume that \begin{align*} \text{Int }X_n = \emptyset \end{align*} for every $n \geq 1$. Then \begin{align*} \text{Int } \Big(\cup_{n=1}^{\infty} X_n\Big) = \emptyset \end{align*}

And I don't understand the first line of the proof. It says,

Set $O_n = X_n^c$, so that $O_n$ is open and dense in $X$ for every $n\geq 1$.

For example, we could let $X = \mathbb{R}$, and $\{X_n\} = [-n,n]$, so that $X = \cup_{1}^{\infty} X_n$, but for any $n$, we have $X_n^c = (-\infty,-n) \cup (n,\infty)$, which is not dense in $\mathbb{R}$. Maybe I am misunderstanding the definition of dense? I thought if $E$ was dense in $F$, then for any point in $x \in F$ and any positive radius $r$, the ball $B(x,r)$ must contain an element of $E$. Certainly we could find a ball in $\mathbb{R}$ that contains no element of $(-\infty,-n) \cup (n,\infty)$, right?

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    Your $X_{n}$ do not have empty interior.:) – 2017-01-05

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Your $X_n$ do not have empty interior, and as the theorem tells you (as $\mathbf R$ is complete), $\mathbf R$ cannot be written as the countable union of closed sets with empty interior (as the union would have empty interior then, and $\mathbf R$ does not).