Proof
Let G be a group of order 6. By Lagrange's Theorem, G has subgroups of order 1,2,3 and 6.
The subgroups of orders 2 and 3 have prime orders and are cyclic therefor. The subgroup contains an element g of order 2 and the subgroup contains an element h of order 3.
So therefore G is cyclic. Is this correct or is there also proof of group G not being cyclic
To answer your question, it is not correct. Several mistakes here.
Lagrange does not say that such subgroups exist. You are incorrectly applying a false converse. Any subgroup of a group of order $6$ must have order $1, 2, 3$, or $6$. This does not say that such subgroups must exist.
Also, a group $G$ of order $6$ having subgroups that are cyclic does NOT mean that the group $G$ itself is cyclic. In fact, all groups have cyclic subgroups by definition (and usually plenty of them).
If you think the statement is false, then your job is to find a group of order $6$ that is not cyclic. If you think it is true then you must prove that all such groups are cyclic. How many groups of order $6$ do you know?
Edit: never mind, answer given away in the comments.