Showing that Lindeberg condition does not hold Let $X_1, X_2, \dots$ be independent random variables and $$X_n = Y_n + Z_n$$ where
$Y_n$ takes values $1$ and $-1$ with chance $1/2$ each, and $$P(Z_n = \pm n) = 1/(2n^2) = (1 - P(Z_n = 0))/2$$ and $S_n:=X_1+ \dots X_n$. Show that Lindeberg condition does not hold, yet $$S_n/\sqrt{n} \rightarrow N(0,1).$$
 A: I will prove more general statement by assuming that $(Y_n)_n$ are i.i.d. such that $\mathbf{E}Y_1=0$ and $\mathbf{E}Y_1^2=1$.
Let $S_n^Y=Y_1+\dots+Y_n$, and $S_n^Z=Z_1+\dots+Z_n$. To show that $S_n/\sqrt{n}\to N(0,1)$ we need two things.  The first thing is that by Lindeberg-Levy theorem (https://en.wikipedia.org/wiki/Central_limit_theorem#Classical_CLT) it holds $S_n^Y/\sqrt{n}\overset{D}{\longrightarrow}N(0,1)$. The second thing is that $S_n^Z/\sqrt{n}\overset{a.s.}{\longrightarrow} 0$ whose proof follows. Namely, since
$$\sum\limits_{n=1}^{\infty} \mathbf{P}(Z_n\ne0)=\sum\limits_{n=1}^{\infty} \frac{1}{n^2}<\infty,$$
Borel-Cantelli lemma implies $\mathbf{P}(Z_n\ne0\text{ i.o.})=0$. This means that $S_n^Z/\sqrt{n}\overset{a.s.}{\longrightarrow} 0$ because for almost every $\omega\in\Omega$ there is $n_0\in\mathbb{N}$ such that for all $n\ge n_0$ it holds $Z_n(\omega)=0$.
Hence, we have proved that $S_n/\sqrt{n}=S_n^Y/\sqrt{n}+S_n^Z/\sqrt{n}\overset{D}{\longrightarrow}N(0,1)$ (using for example Slutsky's theorem).
Now we show that Lindeberg's condition does not hold. I use the following version of the expression inside the condition:
$$
\frac{1}{s_n^2}\sum\limits_{k=1}^n \mathbf{E}[(X_k-m_k)^2;|X_k-m_k|\ge \varepsilon s_k],$$
where $m_k=\mathbf{E}X_k$ and $s_k^2=Var(X_1+\dots+X_k)$, $k\in\mathbb N$. Obviously, in our case $m_k=\mathbf{E}Y_k+\mathbf E Z_k=0$, and because $Var(Z_k)=1$ we have $s_k^2=2k$, $k\in \mathbb N$.
Let's calculate
$\begin{align}
\mathbf{E}[&(X_k-m_k)^2;|X_k-m_k|\ge \varepsilon s_k]= \mathbf{E}[(Y_k+Z_k)^2;|Y_k+Z_k|\ge \varepsilon s_k]\\
&=\mathbf{E}[(Y_k+k)^2;|Y_k+k|\ge \varepsilon s_k,Z_k=k]+ \mathbf{E}[(Y_k+0)^2;|Y_k+0|\ge \varepsilon s_k,Z_k=0]+\mathbf{E}[(Y_k-k)^2;|Y_k-k|\ge \varepsilon s_k,Z_k=-k]\\
&\overset{independece}{\ge} \mathbf{E}[(Y_k+k)^2;|Y_k+k|\ge \varepsilon s_k]\frac{1}{2k^2}+ \mathbf{E}[(Y_k-k)^2;|Y_k-k|\ge \varepsilon s_k]\frac{1}{2k^2}\\
&\overset{iden. distr.}{\ge} \mathbf{E}[(Y_1+k)^2;|Y_1+k|\ge \varepsilon s_k]\frac{1}{2k^2}+\mathbf{E}[(Y_1-k)^2;|Y_1-k|\ge \varepsilon s_k]\frac{1}{2k^2}.\tag{1}
\end{align}$
Fatou's lemma implies $\liminf\limits_{k\to\infty}\mathbf{E}[(Y_1\pm k)^2;|Y_1\pm k|\ge \varepsilon s_k]\frac{1}{2k^2}\ge\mathbf{E}\left[\liminf\limits_{k\to\infty}\left(\frac{(Y_1\pm k)^2}{2k^2}\mathbb{1}_{\{|Y_1\pm k|\ge \varepsilon \sqrt{2k}\}}\right)\right]=\mathbf E[\frac12]=\frac12$. Thus, for sufficiently large $n_0\in\mathbf N$ we have for all $k> n_0$ that $\mathbf{E}[(Y_1\pm k)^2;|Y_1\pm k|\ge \varepsilon s_k]\frac{1}{2k^2}\ge \frac13$. Finally, using $(1)$ we have
$\begin{align}
\liminf_{n\to\infty}\frac{1}{s_n^2}\sum\limits_{k=1}^n \mathbf{E}[(X_k-m_k)^2;|X_k-m_k|\ge \varepsilon s_k]\ge\liminf_{n\to\infty}\frac{1}{2n}\left(\sum\limits_{k=n_0+1}^n \mathbf{E}[(Y_1+k)^2;|Y_1+k|\ge \varepsilon s_k]\frac{1}{2k^2}+\sum\limits_{k=n_0+1}^n \mathbf{E}[(Y_1-k)^2;|Y_1-k|\ge \varepsilon s_k]\frac{1}{2k^2}\right)\ge \liminf_{n\to\infty}\frac{1}{2n}\cdot \left((n-n_0)\cdot\frac13+(n-n_0)\cdot\frac13\right)=\frac13,
\end{align}$
i.e. we have showed that Lindeberg's condition does not hold.
A: It seems that the goal of the example is to give an example where the classical central limit theorem holds  but the Lindeberg condition is not satisfied. Hence the later is sufficient for the central limit theorem but not necessary. 
I will assume that the collection of random variables $\left\{Y_i,Z_j,i,j\geqslant 1\right\}$ is independent. 
By the Borel-Cantelli lemma, there exists $\Omega'$ of probability one such that for all $\omega\in \Omega'$, there exists an integer $N(\omega)$ for which $X_n(\omega)=0$ for all $n\geqslant N(\omega)$ hence $n^{-1/2}\sum_{i=1}^nZ_i\to 0$ almost surely. From the classical central limit theorem applied to the i.i.d. sequence $\left(Y_i\right)_{i\geqslant 1}$ and Slutsky's theorem, it follows that $n^{-1/2}\sum_{i=1}^nX_i\to N(0,1)$.
Observe that $\left(X_i\right)_{i\geqslant 1}$ is independent and that $\mathbb E\left[X_i^2\right]= \mathbb E\left[Y_i^2\right]+\mathbb E\left[Z_i^2\right]=2$
hence Lindeberg's condition reads 
$$
\forall \varepsilon\gt 0,\frac 1n\sum_{i=1}^n
\mathbb E\left[X_i^2\mathbb 1\left\{\left\lvert X_i\right\rvert \gt \varepsilon \sqrt n\right\}\right]\to 0.
$$
Observe that for all $i$, 
$$X_i^2\mathbb 1\left\{\left\lvert X_i\right\rvert \gt \varepsilon \sqrt n\right\}\geqslant \left(i+1\right)^2\left[i+1>\varepsilon\sqrt n\right] \mathbf 1\{Y_i=1\}\mathbf 1\{Z_i=i\}$$
where $[P]$ equal $i$ if assertion $P$ holds and $0$ otherwise. Therefore, 
$$\mathbb E\left[X_i^2\mathbb 1\left\{\left\lvert X_i\right\rvert \gt \varepsilon \sqrt n\right\}\right]\geqslant \left(i+1\right)^2\left[i+1>\varepsilon\sqrt n\right] \Pr\{Y_i=1\}\Pr 1\{Z_i=i\}\geqslant \frac 14\left[i+1>\varepsilon\sqrt n\right].$$
It follows that 
$$
\frac 1n\sum_{i=1}^n
\mathbb E\left[X_i^2\mathbb 1\left\{\left\lvert X_i\right\rvert \gt \varepsilon \sqrt n\right\}\right]\geqslant \frac 1n\sum_{\varepsilon\sqrt n\leqslant i\leqslant n}\frac 14=\frac{n-\varepsilon\sqrt n}{4n}.
$$
A: Partial answer: Calculation that $$Var\left(\frac{S_n}{\sqrt{n}}\right)\neq1$$ Firstly we have that $$E[Y_n]=-1\cdot \frac12 + 1\cdot \frac12=0$$ and  $$E[Y_n^2]=(-1)^2\frac12+(1)^2\frac12=1$$ which gives $$Var(Y_n)=1-0^2=1$$ Similarly we have that $$E[Z_n]=-n\cdot\frac{1}{2n^2}+n\cdot\frac{1}{2n^2}+0=0$$ and $$E[Z_n^2]=(-n)^2\frac{1}{2n^2}+(n)^2\frac{1}{2n^2}+0=1$$ which gives $$Var(Z_n)=1-0^2=1$$ Therefore, assuming independence of $Y_n$ and $Z_n$ we have that $$Var(X_n)=Var(Y_n)+Var(Z_n)=1+1=2$$ which gives that $$Var\left(\frac{S_n}{\sqrt{n}}\right)=\frac1n \sum_{i=1}^{n} Var(X_i)=\frac1n \cdot n2=2 \neq1$$ as you mentioned in the comment. 

Not an answer (just calculations of characteristics functions). We have that $$φ_{\frac{S_n}{\sqrt{n}}}(t)=φ_{X_1}\left(\frac{1}{\sqrt{n}}t\right)\cdotφ_{X_2}\left(\frac{1}{\sqrt{n}}t\right)\cdot \ldots\cdotφ_{X_n}\left(\frac{1}{\sqrt{n}}t\right)$$ where $$φ_{X_k}\left(\frac{1}{\sqrt{n}}t\right)=φ_{Y_k+Z_k}\left(\frac{1}{\sqrt{n}}t\right)=φ_{Y_k}\left(\frac{1}{\sqrt{n}}t\right)\cdotφ_{Z_k}\left(\frac{1}{\sqrt{n}}t\right)$$ with $$φ_{Y_k}\left(\frac{1}{\sqrt{n}}t\right)=E[e^{\frac{it}{\sqrt{n}}Y_k}]=\frac12 e^{-\frac{it}{\sqrt{n}}}+\frac12 e^{\frac{it}{\sqrt{n}}}$$ for all $k \in \mathbb N$ and $$φ_{Z_k}\left(\frac{1}{\sqrt{n}}t\right)=E[e^{\frac{it}{\sqrt{n}}Z_k}]=\frac1{2k^2} e^{-\frac{itk}{\sqrt{n}}}+\frac1{2k^2} e^{\frac{itk}{\sqrt{n}}}+\left(1-\frac1{k^2}\right)\cdot1$$ Thus $$φ_{X_k}\left(\frac{1}{\sqrt{n}}t\right)=\left(\frac12 e^{-\frac{it}{\sqrt{n}}}+\frac12 e^{\frac{it}{\sqrt{n}}} \right)\cdot \left(\frac1{2k^2} e^{-\frac{itk}{\sqrt{n}}}+\frac1{2k^2} e^{\frac{itk}{\sqrt{n}}}+\left(1-\frac1{k^2}\right) \right)=\\=\frac{1}{4k^2}\left(e^{\frac{-it}{\sqrt{n}}(1+k)}+e^{\frac{-it}{\sqrt{n}}(1-k)}+e^{\frac{-it}{\sqrt{n}}(k-1)}+e^{\frac{-it}{\sqrt{n}}(-1-k)}\right)+\frac12\left(1-\frac1{k^2}\right)\left(e^{\frac{-it}{\sqrt{n}}}+e^{\frac{it}{\sqrt{n}}}\right)$$ and $$φ_{\frac{S_n}{\sqrt{n}}}(t)=\prod_{k=1}^{n}\left[\frac{1}{4k^2}\left(e^{\frac{-it}{\sqrt{n}}(1+k)}+e^{\frac{-it}{\sqrt{n}}(1-k)}+e^{\frac{-it}{\sqrt{n}}(k-1)}+e^{\frac{-it}{\sqrt{n}}(-1-k)}\right)+\frac12\left(1-\frac1{k^2}\right)\left(e^{\frac{-it}{\sqrt{n}}}+e^{\frac{it}{\sqrt{n}}}\right) \right]$$ which shows that calculating the characteristic function of $S_n/\sqrt{n}$ is not easy. 

So, you should proceed as in the Proof of CLT presented here. For any random variable, $X$, with zero mean and a unit variance ($\mathrm{Var}(X) = 1$), the characteristic function of $X$ is, by Taylor's theorem, $$\varphi_X(t) = 1 - {t^2 \over 2} + o(t^2), \quad t \rightarrow 0$$ Now, the random variables $\frac{X_k}{\sqrt{2}}$ have the desired property. Thus, although they are not i.i.d. the random variable $$Z_n=\frac{S_n}{\sqrt{2n}}$$ converges to the standard nomral distribution from which you have that  $$\frac{S_n}{\sqrt{n}} \to N(0,2)$$
