Variance of the Euclidean norm under finite moment assumptions Let $X = (X_1,X_2 \cdots X_n)$ be random vector in $R^n$  with independent coordinate $X_i$ that satisfy $E[X_i^2]=1$ and $E[X_i^4] \leq K^4$. Then show that  $$\operatorname{Var}(\| X\|_2) \leq CK^4$$ 
where $C$ is a absolute  constant and $\|   \ \|_2$ denotes euclidian norm.
Here is my attempt:
$$\begin{align*}   E(\|X\|_2^2 -n)^2 &= E[(\sum_{i=1}^n X_i^2)^2 ]-n^2 \\
&=E[\sum_{i=1}^n X_i^4]+E[\sum_{i<j}X_i^2X_j^2]  -n^2 \\
&\leq nK^4 +  2{{n}\choose {2}}-n^2 \\
&\leq n(K^4-1) \\
& \leq nk^4
\end{align*}$$
since $$ E(\|X\|_2^2 -n)^2 \leq nk^4 \rightarrow E\left(\frac{\|X\|_2^2}{n} -1\right)^2 \leq \frac{K^4}{n}$$
and  since 
$$(\forall z \geq 0 \ \ |z-1|\leq |z^2-1|) \rightarrow
  E(\frac{\|X\|_2}{\sqrt n} -1)^2\leq E(\frac{\|X\|_2^2}{n} -1)^2 $$
thus: 
$$E(\frac{\|X\|_2}{\sqrt n} -1)^2 \leq K^4/n  \rightarrow E(\|X\|_2-\sqrt n)^2\leq K^4$$
by Jensen inequality:
$$(E[\|X\|_2] - \sqrt n)^2 \leq K^4 $$
which is equivalence to 
$$ |E[\|X\|_2] - \sqrt n)| \leq K^2$$
then when I am trying to bound $Var(\| X\|_2)$ I meet some problem :
$$\operatorname{Var}(\| X\|_2)=E[\|X\|_2^2] -(E[\|X\|_2])^2 \leq n- (K^2-\sqrt n)^2 \leq -K^4+2K^2\sqrt n$$ which is not bound by constant , how can I bound that?

You had most of the steps correct. 
As you argued correctly, $$ E (\|X\|_2 - \sqrt{n})^2 \leq K^4 . $$
Note that the mean minimizes the squared error, i.e., for any $c \in \mathbb{R}$, $Var(X) \leq E(X-c)^2$. Therefore, 
$$Var(\|X\|_2) \leq E (\|X\|_2 - \sqrt{n})^2 \leq K^4 .$$
