Let n = 2r, and n = xx for some integers r and x, because n is even and n is a square. So xx = 2r.
Because of the fundamental theorem of arithmetic, we know that x must be representable as the product of a unique string of prime numbers.
Because 2 is prime, then since xx = 2r, there must be a 2 in the string of primes for xx.
But since 2 is prime, it must be in x as well, because a prime cannot come out of nowhere. In other words, if there is a given prime P in xx, there must be at least two P in xx, because there was at least one in x, and the number of each one got doubled in xx.
Therefore xx = 2r = 2*2*y = 4y for some integer y.
Therefore n = 4y and sqrt(n) = sqrt(4y) = sqrt(4)sqrt(y) = 2sqrt(y) which is an even number.
FTA is massive overkill. For every number n, either n can be expressed as 2k for some k, or 2k+1 for some k, but not both (proof: by induction); in particular the square root can too. If the square root is (2k+1), then the square is 4k^2 + 4k + 1 = 2(2k^2+2k) + 1, which is by definition odd, not even as we supposed.
The FTA proof is the one that's obvious, though. If it's really easy to do something using a basic tool, why worry that the basic tool is complex to describe?
The proof would be more compelling if this was proven instead of being taken as an obvious fact.