Let $X$ be a non-empty compact subset of the plane and $\{Y_{\alpha}\}$ a family of pairwise disjoint compact subsets of the plane each of which intersects $X$ in a singleton. Is it possible to choose $X$ and $\{Y_{\alpha}\}$ such that $X\cup\bigcup_{\alpha}Y_{\alpha}$ covers the plane?
If the compactness constraint on the sets $Y_{\alpha}$ were dropped then a solution would be to take $X$ as the closed unit disc and the sets $Y_{\alpha}$ as the spokes radiating from the boundary of the disc.
But in my case, the sets $\{Y_\alpha\}$ have to be compact. I suspect that I am asking the impossible of $X$ and $\{Y_{\alpha}\}$ but I have not been able to prove this.
Here's a construction for the special case where the compact set $X \subset \mathbb R^2$ has the same cardinality as $\mathbb R$, for example if $X$ contains an open ball. We can take the index set $\{\alpha\}$ to be $X$ itself (because your requirements yield one subset $Y_\alpha$ for each $\alpha \in X$).
Each $Y_\alpha$ will be a two point set, hence compact: the point $\alpha \in X$, and one other point $\beta(\alpha) \in \mathbb R^2 - X$ depending on $\alpha$.
I'll use the fact that the sets $X$ and $\mathbb R^2 - X$ are both in one-to-one correspondence with $\mathbb R$ and hence in one-to-one correspondence with each other. Choosing a bijection $\beta : X \to \mathbb R^2 - X$, as $\alpha \in X$ varies the sets $Y_\alpha = \{\alpha,\beta(\alpha)\}$ satisfy your requirements.