Answer
Mirsky's theorem (1971) in order theory has been discovered late.
Background
In 1950, Robert Dilworth published what is today known as Dilworth's theorem:
In any finite poset $(X,\leq)$, the size of the largest antichain equals the minimum number of blocks of a partition of $X$ into chains.
By now, several proofs are known, but for the "hard" implication "$\Rightarrow$", all of them are somewhat involved inductive proofs.
There is a pretty natural "dual" of the statement, today known as Mirsky's theorem:
In any finite poset $(X,\leq)$, the size of the largest chain equals the minimum number of blocks of a partition of $X$ into antichains.
Surprisingly, apparently no one had thought about that dual statement for about two decades, until it was discovered and published in 1971 by Leon Mirsky. All the more as the proof turned out to be much simpler than everything known for Dilworth's theorem. For the hard implication "$\Rightarrow$", there is a direct construction of the partition as the preimages of the map
$$X \mapsto \mathbb{N},\quad x\mapsto \text{size of the largest chain with maximal element }x.$$
Nothing comparable is known for Dilworth's theorem.
More details on Mirsky's 1971 publication
Mirsky's theorem was published in the article [L. Mirsky: A dual of Dilworth's decomposition theorem, The American Mathematical Monthly 78[8] (1971) 876-877.]
From the article:
We owe to Dilworth [1] the following well-known and important decomposition theorem:
Theorem 1 [...]
It may be of some interest to note that this statement remains valid if the roles of chains and antichains are interchanged. Thus we have the following result:
Theorem 2 [...]
Thus, in a formal sense, Theorem 2 may be regarded as a 'dual' of Theorem 1. However, as we shall see, the proof of the dual result is considerably easier than that of Dilworth's original theorem. [...]