Question: A finite Hausdorff topological group, has discrete topology and every discrete group is totally disconnected. I look for an example of a non-Abelian, finite, connected non-Hausdorff group . I think every non-Abelian, finite, connected groups must be discrete.

Answer: 
Here is one approach.

Let $b > 1$. If you compute the derivative of the function $b^x$, you find that the answer is just $b^x$ multiplied by an (annoying) constant.

There is a value of $b$ for which this constant is equal to $1$. That's nice! With this special value of $b$, the derivative of $b^x$ is just $b^x$, the same thing we started with. That's a very neat property for a function to have.

This special value of $b$ is $e = 2.718 \ldots$.

It is now easy to compute the Taylor series of the function $e^x$ (centered at $0$). We find that
\begin{equation}
\tag{$\spadesuit$} e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots.
\end{equation}
This comes directly from the Taylor series formula
\begin{equation}
f(x) = f(x\_0) + f'(x\_0)(x - x\_0) + \frac{f''(x\_0)}{2!}(x - x\_0)^2 + \cdots.
\end{equation}

Plugging $x = 1$ into ($\spadesuit$) yields
\begin{equation}
e = \sum\_{n=0}^{\infty} \frac{1}{n!}.
\end{equation}
Let $x \in \mathbb{R}^+$.

$f:x\longrightarrow \exp(x) \in C^\infty(\mathbb{R}, \mathbb{R})$, hence we can write $\forall n \in \mathbb{N}$ :

$$\left|f(x)-\sum\_{k=0}^n f^{(k)}(0)\frac{x^k}{k!}\right| \leq \frac{|x|^{n+1}}{(n+1)!}I\_{n+1} \text{ where } I\_{n+1}=sup\_{[0,x]}|f^{n+1}|$$

And $\forall t \in [O,x] f^{(n)}(t)=\exp(t)$ which means $I\_{n+1}=\exp(x)$

But $\displaystyle\lim \_{n\to \infty }\left(\frac{|x|^{n+1}}{(n+1)!}\exp(x)\right)=\:0$

So $\displaystyle\lim \_{n\to \infty }\left(\sum\_{k=0}^n\frac{x^{k}}{k!}\right)=\exp(x)$

Just take $x = 1$ for $\mathbb{e}$