We say that a metric space $X$ is a Baire space if there is no open set $E$ such that $$E \subseteq \bigcup\limits_{n\geq 1} F_i,$$ in which each $F_i$ is a closed set with empty interior.
Suppose that $X$ has the following property: if $X=\bigcup\limits_{n\geq 1} F_i$, in which each $F_i$ is a closed set, then $\bigcup\limits_{n\geq 1} \operatorname{int}(F_i)$ is dense in $X$, in which $\operatorname{int}(F_i)$ denotes the interior of $F_i$. Prove that $X$ is a Baire space.
My attempts so far consist in supposing that there is an open set contained in a union of closed sets with empty interior, and trying to obtain a union of closed sets which is the whole space, so I can use the hypothesis, but I'm stuck there. Any hints?
In the first line you should say that $E$ is a non-empty open set.
$E \subseteq \bigcup\limits_{n\geq 1} F_i$ ($E$ non-empty) implies that $X= \bigcup\limits_{n\geq 1} F_i \cup E^{c}$. Since $F_i$ has no interior for each $i$ it follows that the interior of $E^{c}$ is dense. But $E$ is an open set disjoint from $E^{c}$, hence also from its interior, so we have arrived at a contradiction. [A dense set has to intersect every non-empty open set].