Proving $\frac{\sqrt{n}(S_n^2-\sigma^2)}{\sqrt{u_2-\sigma^4}}\rightarrow N(0,1)$ in distribution 
Let $X_l, X_2,\ldots$ be i.i.d. with $E(X_1 ) = u$, $\operatorname{Var } (X_1) = \sigma^2$ and  $u_2 = E[(X_1 - u)^4]$. Let $S_n^2 = \frac{1}{n} \sum_{i=1}^{n} (X_i - \bar{X}_n)^2$ where $\bar{X}_n = \frac{\sum_{n=1}^{n} X_i}{n}$. Prove that $\frac{\sqrt{n}(S_n^2-\sigma^2)}{\sqrt{u_2-\sigma^4}}\rightarrow N(0,1)$ in distribution sense.

My thought: I tried to prove this using Levy-Cramer theorem with the help of Taylor series expansion, but I could not find the characteristic function of the given messy expression. Could someone please help find $\phi_{\frac{\sqrt{n}(S_n^2-\sigma^2)}{\sqrt{u_2-\sigma^4}}} (t)$ in terms of $\phi_{X} (h(t))$ where $h(t)$ could be any functions as long as the transformation makes sense? Otherwise, any inputs would really be appreciated.
 A: We can replace $X_n$ by $X_n-u$, hence we can assume that $u=0$. We haved 
\begin{align}
 \sqrt n\left( S_n^2-\sigma^2\right)&=\frac 1{\sqrt n}\sum_{i=1}^n \left(X_i^2 -\sigma^2\right) -2 \sqrt n\overline{X_n}^2 +  \sqrt n\overline{X_n}^2\\
&=\frac 1{\sqrt n}\sum_{i=1}^n \left(X_i^2 -\sigma^2\right) - \sqrt n\overline{X_n}^2.
\end{align}
Now, notice that $$\tag{*}   \sqrt n\overline{X_n}^2 =\frac 1{\sqrt n}\left(\frac 1{\sqrt n}  \sum_{i=1}^nX_i   \right)^2, $$
hence by Slutsky's lemma and the central limit theorem, $\sqrt n\overline{X_n}^2\to 0$ in probability. 
In order to reach the wanted conclusion, use the central limit theorem for the i.i.d. centered sequence $\left(\left(X_i^2-\sigma^2\right)/\sqrt{ u_2-\sigma^2} \right)_{i\geqslant 1}$ (because the variance of $X_1^2-\sigma^2$ is $\mathbb E\left[X_1^4\right]-\sigma^4=u_2-\sigma^4$). 
A: First assume that $u = 0$ without loss of generality. You can do this because $S_n^2$ doesn't change if you consider $X_i-u$ instead of $X_i$.
We are going to need the multivariate central limit theorem here. It is the same as the ordinary central limit theorem but instead of variance you have the covariance matrix of the random vector. Note that the sequence $\left(\begin{pmatrix}X_i\\ X_i^2\end{pmatrix}\right)_i$ is i.i.d and it does have a covariance matrix since the fourth moment of $X_1$ exists as stated in your question. So let's apply the central limit theorem to this sequence.
$$\sqrt{n}\left(\begin{pmatrix}\frac{1}{n}\sum_{i=1}^nX_i\\ \frac{1}{n}\sum_{i=1}^nX_i^2\end{pmatrix} - \begin{pmatrix}0\\ \sigma^2 \end{pmatrix}\right) \rightsquigarrow N(0,\Sigma)$$
Here $\Sigma$ denotes the covariance matrix of  $\begin{pmatrix}X_1\\ X_1^2\end{pmatrix}$.
Next we are going to use the delta method. Note that $S_n^2 = \frac{1}{n}\sum_{i=1}^nX_i^2 - \bar{X}_n^2$. So then (using the notation on the Wikipedia page) the transformation we have is $g(x,y) = y - x^2$. Make sure you check the conditions needed on $g$ for the delta method to work.
So then,
$$\sqrt{n}\left(S_n^2-\sigma^2\right) \rightsquigarrow N(0,\begin{pmatrix}0 & 1\end{pmatrix}\Sigma\begin{pmatrix}0 \\ 1\end{pmatrix})$$
By construction $\begin{pmatrix}0 & 1\end{pmatrix}\Sigma\begin{pmatrix}0 \\ 1\end{pmatrix} = Var(X_1^2)$. Check that $Var(X_1^2) = u_2 - \sigma^4$.
Hence
$$\frac{1}{\sqrt{u_2 - \sigma^4}}\sqrt{n}\left(S_n^2-\sigma^2\right) \rightsquigarrow N(0,1)$$
