Non-abelian finite group $G$ has at least 2 conjugacy classes which contain at least 2 elements 
Show that each non-abelian finite group $G$ has at least $2$ conjugacy classes which contain at least $2$ elements.

I have a solution which uses the class equation. However, as we have not dealt with this equation in class, we do not have it at our disposal.

Is it possible to prove this proposition without the class equation? 

My attempt looks like this:
As $G$ is non-abelian, there are $a$ and $b$ with $a\neq b$ such that $aba^{-1}\neq b $  and also $bab^{-1}\neq a $. Hence, in the conjugacy class of $b$, there are $b$ itself and $aba^{-1}$, which is why this class contains at least two distinct elements. The same can be done for the conjugacy class of $a$ which contains at least $a$ and $bab^{-1}$. At first glance, both classes seem to be distinct. 

However, could it be possible that both conjugacy classes are equal? 

Then, this attempt of a proof would be invalid, of course. 
 A: To arrive at a contradiction assume for the moment that $G$ has only a single conjugacy class of cardinality greater than $1$. This implies that all non-central elements form a conjugacy class, say $G-Z(G)=Cl_G(x)$ for some $x \notin Z(G)$. Note, since $x\neq x^2$, $|C_G(x)| \geq 2$. This yields $|G:C_G(x)|=\#Cl_G(x) \leq |G|/2$, whence $|G|-|Z(G)| \leq |G|/2$, being equivalent to $|G|/2 \leq |Z(G)|$. But $G$ is not abelian, that is, $|Z(G)| \lt |G|$ and we conclude $|G|/2=|Z(G)|$ so $G/Z(G)$ is cyclic (of order $2$), implying $G$ is abelian, a contradiction!
Note (added January 19th 2022) Here is a somewhat different and streamlined proof.
Proposition There do not exist finite groups $G$ having a single conjugacy class of cardinality $\gt 1$.
Proof Let $n \gt 1$ be the cardinality of the conjugacy class. Then by the Class Formula we have
$$|G|=|Z(G)|+n$$
Since $n$ is the index of some centralizer, by taking the equation mod $n$ we see that $|Z(G)| \equiv 0$ mod $n$, that is, $n \mid |Z(G)|$. Taking the equation mod $|Z(G)|$, we also see that $|Z(G)| \mid n$. We conclude that $|Z(G)|=n$ and $|G|=2n$. So, $|G/Z(G)| \cong C_2$ is cyclic and $G$ must be abelian, contradicting $n \gt 1$. $\square$
This line of reasoning also generalizes nicely to the non-existence of finite groups $G$ having exactly 2 conjugacy classes of the same cardinality.
A: The beginning of your reasoning is correct.
Suppose, for a contradiction, that there is only one conjugacy class with more than $1$ element.  Then every non-central element is conjugate to every other non-central element.
In case all elements of $G$ are of order 2, then for any element $a,c\in G$ you will have $cac^{-1}=cac=caca\cdot a^{-1}=a^{-1}=a$.  So it is quite easy to analyze what will happen in this case.
Otherwise, pick an element $a\in G$ of order more than $2$.  Then there are $|G\setminus Z(G)|-o(a)+1$ elements to conjugate $a$ by (since conjugating $a$ by its powers or by things in the center will result in just $a$ again).  Therefore, there must be at least $o(a)-1+|Z(G)|$ elements that $a$ is not conjugate to.
Obviously, this would not be the case if $G$ were infinite.
A: You can find an element $a \in G$ that does not commute with $b \in G$. As you showed, this implies that the class $Cl_G(a)$ has cardinality at least $2$, or, equivalently, the index of the centralizer of $a$, $|G:C_G(a)| \geq 2$. This makes $C_G(a)$ a proper subgroup. Now use the fact that in a finite group, the union of all the conjugates of a proper subgroup cannot be the whole group (see for example here). It follows that you can find an $x \in G$, with $x \notin \bigcup_{g \in G}C_G(a)^g=\bigcup_{g \in G}C_G(a^g)$. This implies that $a$ and $x$ do not commute and also that $Cl_G(a) \cap Cl_G(x) = \emptyset$.
