= Solution
König's theorem for <cardinal numbers> states that, for a set-indexed family with $\kappa_i<\lambda_i$ at every index,
$$
\boxed{\sum_{i\in I}\kappa_i<\prod_{i\in I}\lambda_i.}
$$
Here the sum is the <cardinality> of a <disjoint union> and the product is the <cardinality> of a <Cartesian product>. Work in <ZFC>. Choose disjoint <sets> $A_i$ of sizes $\kappa_i$ and <sets> $B_i$ of sizes $\lambda_i$. For any <function> $f:\bigsqcup_i A_i\to\prod_iB_i$, its $i$th coordinate values on $A_i$ form a <subset> of $B_i$ of <cardinality> at most $\kappa_i$. Choose $b_i\in B_i$ outside that <subset>, using <choice>. Then $b=(b_i)_i$ cannot equal $f(a)$ for any $a$: if $a\in A_j$, its $j$th coordinate was specifically excluded. Thus there is no <surjection> from the <disjoint union> onto the product.
<cardinal number>[Cardinal comparability] under <choice> now implies the displayed strict inequality. Empty index <set> gives $0<1$ directly; empty individual $A_i$ cause no difficulty, since the corresponding coordinate excludes an empty image. This is the cardinal diagonal theorem, rather than the finitely branching tree result also bearing König's name.
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