Colour the countable subsets of a cardinal so every $\kappa$-sized set is polychromatic (Erdős #598)

Let $m$ be an infinite cardinal and let $\kappa=(2^{\aleph_0})^{+}$ be the successor cardinal of the continuum. Can one colour the countable subsets of $m$ — that is, all subsets of $m$ of size $\aleph_0$ — using $\kappa$ many colours in such a way that every $X\subseteq m$ with $\lvert X\rvert=\kappa$ contains, among its countable subsets, subsets of all $\kappa$ colours? Equivalently: is there a colouring $c:[m]^{\aleph_0}\to\kappa$ such that every $\kappa$-sized $X\subseteq m$ is 'polychromatic', meeting all $\kappa$ colour classes?

Acceptance. FULLY RESOLVES: a complete proof (machine-checkable in Lean/Coq preferred, given the formalised statement) either constructing, in ZFC, a colouring $c:[m]^{\aleph_0}\to\kappa$ with $\kappa=(2^{\aleph_0})^{+}$ for which every $\kappa$-sized $X\subseteq m$ is polychromatic, or proving no such colouring exists — either for all infinite $m$, or with the dependence on $m$ made precise. ADVANCES: settle the question for a significant special case (for example the least non-trivial value of $m$, or under a stated hypothesis such as CH or a fixed value of $2^{\aleph_0}$) with full proof; or a consistency / independence result covering some parameter regime, with proof. Deliver the construction / proof or the forcing argument with verification.