Definition
A subset $Y$ of a topological space $X$ is said nowhere dense if $\operatorname{int}(\operatorname{cl} Y)$ is empty.
Now let be $X$ a compact topological space and thus let be
$$
\mathcal Y:=\{Y_n:n\in\omega\}
$$
a countable collection of nowhere dense subsets of $X$. So I am searching to prove that $X\setminus\bigcup\cal Y$ is dense in $X$ using the following hint given by my topology text.
Hint: So given an open subset $A$ we suppose that
$$
\mathcal A:=\{A_n:n\in\omega\}
$$
is a countable collection of nonempty open sets such that
$$
A_1\subseteq\operatorname{cl}A_1\subseteq A
$$
and in particular such that
$$
\operatorname{cl}A_{n+1}\subseteq A_n\quad\text{and}\quad A_n\cap Y_n=\emptyset
$$
for each $n\in\omega$
So we observe that the inclusion
$$
\operatorname{cl}A_{n+1}\subseteq A_n\subseteq\operatorname{cl}A_n
$$
proves that the collection
$$
\overline{\mathcal A}:=\{\operatorname{cl}A_n:n\in\omega\}
$$
is a collection of closed sets with the finite intersection property so that by compactness $\bigcap\overline{\mathcal A}$ is a nonempty subset of $A$ and thus in particular also
$$
\bigcap_{n\in\omega}\operatorname{cl}A_{n+1}
$$
is a nonempty subset of $A$.
Now we observe that the if $Y_n$ and $A_n$ are disjoint for each $n\in\omega$ then the inclusion
$$
\operatorname{cl} A_{n+1}\subseteq A_n\subseteq X\setminus Y_n
$$
holds for any $n\in\omega$ and thus we conclude that also the inclusion
$$
\bigcap_{n\in\omega}\operatorname{cl}A_{n+1}\subseteq\bigcap_{n\in\omega}X\setminus Y_n=X\setminus\bigcup_{n\in\omega} Y_n
$$
holds and this proves that $A$ is not disjoint from $X\setminus\bigcup\mathcal Y$ so that this is dense.
So I tried to make the collection $\cal A$ as follows. First of all I observe that a finite union of nowhere dense sets is nowhere dense so that $\bigcup_{i\in n} Y_i$ is nowhere dense for each $n\in\omega$ and thus $A$ is not contained in $\bigcup_{i\in n}\operatorname{cl} Y_i$ so that
$$
A\cap\biggl(\bigcap_{i\in n}(X\setminus \operatorname{cl}Y_i)\biggl)=A\cap\biggl(X\setminus\bigcup_{i\in n}\operatorname{cl}Y_i\biggl)\neq\emptyset
$$
Therefore I thought to put
$$
A_n:=A\cap\biggl(\bigcap_{i\in n}(X\setminus \operatorname{cl}Y_i)\biggl)
$$
for each $n\in\omega$ but unfortunately I was not able to prove that the last position works so that I thought to put a specific question. So could someone help me, please?
Best Answer
I might be wrong, in which case I'll delete this answer, but I think your problem is that your assumption of compactness is not enough. For example, take the one-point compactification of the rationals $\mathbb{Q}^*$: it is the countable union of the singletons $\{x\}$ for $\mathbb{Q}\cup \{\infty\}$, each of which are nowhere dense, so it is compact and yet not Baire.
EDIT: Now that I've thought about it a little more I additionally found this counterexample: $\mathbb{Q}$ under the cofinite topology is compact and notably fails horribly to be Hausdorff, but every singleton $\{x\}$ for $x\in \mathbb{Q}$ is closed with empty interior so it is a countable union of nowhere dense sets.
To sort out your problem, my best guess is that you meant to include the additional assumption of Hausdorffness because this implies your space is regular and then you can construct your $A_i$ sets as described here (there, they refer to what you call $A_i$ as $O_i$ and refer to your $Y_i$ as $A_i$).