I will proove it using contradiction. Assuming no element has order 2. i.e. $a^2 \neq e$ for any non-identity element a, means
$a \neq a^{-1}$ for any (non-identity element) a.
Further rewriting meaning: the inverse of any element is not that element itself, it is someone else.
But i want to somehow show that at least one element has inverse as its own. I am trying a method lets see if it works :).
i will select each element from the set and will check inverse of each element, and in this process, as soon as i encounter any element having $a^2 \neq e$ then i am done.
My Goal: to show G has atleast 1 element as its own inverse.
Let $|G| = 2n$. Then we take out the identity and have 2n-1 elements to choose from.
Step1: select an element, if it is its own inverse then i am done.
Step2 (otherwise) : if inverse of a is not a and its b. then throw a and b out. (bcoz if 'a' inverse is 'b' then 'b' inverse also 'a', and inverse of an element is unique.)
(Notice we always throw one pair)
In worst case i will ended up throwing each element but one element will left as it is, bcoz total number of elements are odd (2n-1) and we always throw a pair of two (even).
Now question is, What is inverse of that element ?
it has to be its own inverse, it can not map to inverse of any other element bcoz inverse is unique. And moreover it can not be inverse of identity element bcoz inverse of identity is identity itself.
Finally, i can say there exist one non-identity element 'a' of order 2.
Yes, it worked !
Hence Proved !