**Counterexample.**

Let $\{\alpha:\alpha\lt\mathfrak c\}=I\cup J$ where $I\cap J=\emptyset,\ |I|=|J|=\mathfrak c.$

Let $A=\{t_\alpha:\alpha\in J\}$ be the Cantor ternary set; $t_\alpha\ne t_\beta$ for $\alpha\ne\beta$.

Let $S$ be a dense $G_\delta$-subset of $[0,1]$ which has Lebesgue measure zero and is disjoint from $A;$ then $T=[0,1]\setminus S$ is a first category subset of $[0,1].$

For every interval $(a,b)\subseteq[0,1]$ we have $|S\cap(a,b)|=|(T\setminus A)\cap(a,b)|=\mathfrak c.$

Let $\{S_\alpha:\alpha\in I\}$ be a partition of $S$ into countable dense sets.

Let $\{T_\alpha:\alpha\in J\}$ be a partition of $T$ into countable dense sets such that $T_\alpha\cap A=\{t_\alpha\}.$

For $\alpha\lt\mathfrak c$ define

$$X_\alpha=\begin{cases}
S_\alpha\ \text{ if }\ \alpha\in I,\\
T_\alpha\ \text{ if }\ \alpha\in J.
\end{cases}$$

$X=[0,1]$ is a compact metric space, and $\{X_\alpha:\alpha\lt\mathfrak c\}$ is a partition of $X$ into $\mathfrak c$ countable dense subsets.

$A$ is a nonempty closed subset of $X$ such that $X_\alpha\cap A=\emptyset$ for $\alpha\in I$ and $X_\alpha\cap A=\{t_\alpha\}$ for $\alpha\in J,$ so that $X_\alpha\cap A$ is first category in $A$ for each $\alpha\lt\mathfrak c.$

Finally, $\bigcup\{X_\alpha:X_\alpha\cap A\ne\emptyset\}=\bigcup\{X_\alpha:\alpha\in J\}=T$ is first category in $X.$