I am trying to prove that $$\frac{S_n=\sum_{k=1}^n X_k^2-n}{\sqrt{n}}$$ converges in distribution, where $X_i$ are iid. and are normally distributed with mean 0 and variance 1.
I defined $X_{n,k}=\frac{X_k^2-1}{\sqrt{n}}$ its expected value is zero, the sum of variances of $X_{n,k}$ converges to 3, so the only thing left is to check Linderberg condition.
$$ \forall\epsilon_{>0}\space lim_{n\to\infty}\sum_{k=1}^{n}|\frac{X_k^2-1}{\sqrt{n}}|1_{|\frac{X_k^2-1}{\sqrt{n}}|>\epsilon}=0$$, but I am not sure on how to do that, the expected value of $X_k^2$ is 1 , so it might be true, but I am not sure how to argument it, as $X_k^2$ is surely not bounded.
The result that you desire is actually true in slightly more generality. Instead of having $X_i \sim^\text{iid} N(0,1)$ -- which we note has fourth moment $E(X^4) = 3$ (see this Wiki page) -- let them be general iid random variables with second moment $E(X^2) = \mu$ and fourth moment $E(X^4) = \sigma^2$.
Now define $Y_i = X_i^2$; note that $E(Y_i) = E(X_i^2) = \mu$ and $E(Y_i^2) = E(X_i^4) = \sigma^2$, and that the $Y_i$ are iid. We may then apply the CLT (see Section 21 of these lecture notes by James Norris for a rigorous, readable treatment of the CLT). Doing this, we deduce that $$S_n = \frac{\sum_{i=1}^n Y_i - n \mu}{\sigma \sqrt n} = \frac{\sum_{i=1}^n X_i^2 - n \mu}{\sigma \sqrt n} \to^d N(0,1).$$
In your specific case, we have $\mu = E(Y_i) = E(X_i^2) = 1$ and $\sigma^2 = E(Y_i^2) = E(X_i^4) = 3$. Hence we deduce that $$ \frac{\sum_{i=1}^n X_i^2 - n}{\sqrt{3n}}.$$ (Note that this is not quite what you thought: you just missed a factor $1/\sqrt3$, due to the fact that $E(X_i^4) = 3$, not $1$.)
I don't know what the Lindenberg condition is (although I recognise the form---I think it's to do with proving Berry-Esséen estimates, which control the error in the approximation); but it is unnecessary for the standard CLT.