Proof: By Euclid

For let $BC$ have been produced to $D$.
And since the angle $ACD$ is external to triangle $ABC$, it is greater than the internal and opposite angle $ABC$ [Prop. 1.16].
Let $ACB$ have been added to both.
Thus, the (sum of the angles) $ACD$ and $ACB$ is greater than the (sum of the angles) $ABC$ and $BCA$.
But, (the sum of) $ACD$ and $ACB$ is equal to two right angles [Prop. 1.13].
Thus, (the sum of) $ABC$ and $BCA$ is less than two right angles.
Similarly, we can show that (the sum of) $BAC$ and $ACB$ is also less than two right angles, and further (that the sum of) $CAB$ and $ABC$ (is less than two right angles).
Thus, for any triangle, (the sum of) two angles taken together in any (possible way) is less than two right angles.
(Which is) the very thing it was required to show.
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