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) $a.$

Rewriting the statement: the inverse of any element is not that element itself, it is something 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 at least $1$ element as its own inverse.

Let $|G| = 2n$. Then we take out the identity and have $2n-1$ elements to choose from.

Step 1: Select an element, if it is its own inverse then I am done.

Step 2 (otherwise): If inverse of $a$ is not $a$ and is $b,$ throw $a$ and $b$ out. $(\because$ if $a$ inverse is $b$ then $b$ inverse is $a$, and inverse of an element is unique$)$

(Notice we always throw one pair)

In worst case I will end up throwing all but one element, because total number of elements is odd $(2n-1)$ and we always throw a pair of two (even).

Now question is, what is the inverse of that element?

**It has to be its own inverse**, it can not map to inverse of any other element because inverse is unique. And moreover it can not be inverse of identity element because 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 !