As K.P. and Ramiro both point out in the comments, it follows from $\mathsf{CH}$ that the answer is no. I claim that it is also consistent that the answer is yes.
It is consistent that $\mathfrak{c} > \aleph_2$ and that there is a partition $\mathcal P$ of the Cantor space $2^\omega$ into $\aleph_2$ closed sets, such that $\aleph_2$ members of $\mathcal P$ are singletons. (See Theorem 3.11 in this paper for a proof.)
Let $Y$ be any completely metrizable space of weight $\kappa = \aleph_1$, and let $X$ be the disjoint sum of $Y$ and the Cantor space. Let $F_0 = Y$, and let $\{F_\alpha :\, 1 \leq \alpha < \omega_2 \}$ be an enumeration of the members of $\mathcal P$ that are not singletons.