Proof: By Euclid

For let the least number measured by $A$, $B$, $C$ have been taken, and let it be $DE$ [Prop. 7.36].
And let the unit $DF$ have been added to $DE$.
So $EF$ is either prime, or not.
Let it, first of all, be prime.
Thus, the (set of) prime numbers $A$, $B$, $C$, $EF$, (which is) more numerous than $A$, $B$, $C$, has been found.
And so let $EF$ not be prime.
Thus, it is measured by some prime number [Prop. 7.31].
Let it be measured by the prime (number) $G$.
I say that $G$ is not the same as any of $A$, $B$, $C$.
For, if possible, let it be (the same).
And $A$, $B$, $C$ (all) measure $DE$.
Thus, $G$ will also measure $DE$.
And it also measures $EF$.
(So) $G$ will also measure the remainder, unit $DF$, (despite) being a number [Prop. 7.28].
The very thing (is) absurd.
Thus, $G$ is not the same as one of $A$, $B$, $C$.
And it was assumed (to be) prime.
Thus, the (set of) prime numbers $A$, $B$, $C$, $G$, (which is) more numerous than the assigned multitude (of prime numbers), $A$, $B$, $C$, has been found.
(Which is) the very thing it was required to show.
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