A standard and rigorous proof using the pigeonhole principle

The Pigeonhole Principle is really the statement that for finite sets $S,T$ with $|S|>|T|$, there is no injective mapping from $S$ to $T$. Proceed by induction on $|T|$. For $|T|=1$, $|S|\ge 2$ by hypothesis; there is one map from $S$ to $T$ and it is not one-to-one. Now suppose true for sets with $n$ elements and that $|T|=n+1$ (and $|S|>n+1$), and let $\phi:S\to T$. Let $t\in T$. Then as there are no injective mappings from $S$ to $T-\{t\}$ by the induction hypothesis, $\phi$ won't be injective unless some $s\in S$ maps to $t$, so suppose this is so. If another element also maps to $T$, then clearly $\phi$ is not one-to-one, so suppose that $\phi^{-1}(t) = \{s\}$. Then $\phi|_{S-\{s\}}$ maps $S-\{s\}$ to $T-\{t\}$, and by the induction hypothesis this map can't be one-to-one. So in all cases $\phi$ fails to be one-to-one.

The pigeonhole principle says that if $i$ items are to be placed into $c$ containers, where $i>c$, then at least one container must have at least two items. The sums of the 20 numbers stand for 20 items and the number of possible results, 19, stands for the containers.

I don't like quoting the pigeonhole principle when it is nothing more than common sense. Like this:

There are only $19$ possible two-digit sums, because the smallest is $0+0$ and the largest is $9+9$ and there are only $19$ integers from $0$ to $18$. So obviously if you have $20$ two-digit numbers, they cannot all have different digit sums, because $20 > 19$.

More generally, if you have $c·k+1$ items and assign one of $c$ labels to each item, there is a label assigned to at least $k+1$ items. Proof: If the claim is false, then every label is assigned to at most $k$ items, so there can be at most $c·k$ items, contradicting the given number of items. Therefore the claim is true.