Is finite and strictly positive variances all that is needed for the CLT Lindeberg condition? I cannot understand what I am doing wrong here, so please somebody, point it out from me.  
The issue? I keep finding that the Lindeberg sufficient condition for the Central Limit Theorem for independent non-identically distributed random variables, holds  already by the premises (i.e. holds always given the premises). The condition is
$$\lim_{n \to \infty} \frac{1}{s_n^2}\sum_{k = 1}^{n} \mathbb{E}\big[(X_k - \mu_k)^2 \cdot \mathbf{1}_{\{ | X_k - \mu_k | > \varepsilon s_n \}}  \big] = 0,\;\;\; 
\forall \varepsilon > 0$$
where $s_n^2 \equiv \sum_{k=1}^n \sigma_k^2$,  
and I keep finding that as long  as
$$0<c\leq\sigma^2_k< \infty,\;\;\; \forall\,k$$
which is part of the premises, the condition will hold. We have
$$\mathbb{E}\big[(X_k - \mu_k)^2 \cdot \mathbf{1}_{\{ | X_k - \mu_k | > \varepsilon s_n \}}  \big]  \\= \mathbb{E}\left[(X_k - \mu_k)^2\mid \{| X_k - \mu_k | > \varepsilon s_n\}\right] \cdot \text{Prob}\left(| X_k - \mu_k | > \varepsilon s_n\right)$$
$$ \leq \sigma^2_k \cdot \text{Prob}\left(\frac{|X_k - \mu_k|}{\sigma_k}> \varepsilon s_n/\sigma_k\right)  $$
because conditional variance is not greater than unconditional variance 
$$\leq\sigma^2_k \cdot \frac {\sigma^2_k}{\varepsilon^2 s_n^2}= \frac {(\sigma^2_k)^2}{\varepsilon^2 s_n^2}$$
by the basic Chebychev inequality. Then 
$$\lim_{n \to \infty} \frac{1}{s_n^2}\sum_{k = 1}^{n} \mathbb{E}\big[(X_k - \mu_k)^2 \cdot \mathbf{1}_{\{ | X_k - \mu_k | > \varepsilon s_n \}} \leq \lim_{n \to \infty} \frac{1}{s_n^2}\sum_{k = 1}^{n} \frac {(\sigma^2_k)^2}{\varepsilon^2 s_n^2}$$ 
$$=\frac 1{\varepsilon^2 } \cdot \lim_{n \to \infty} \frac{\sum_{k = 1}^{n}(\sigma^2_k)^2}{\left(\sum_{k=1}^n \sigma_k^2\right)^2} =  \frac 1{\varepsilon^2 } \cdot \lim_{n \to \infty} \frac{\sum_{k = 1}^{n}(\sigma^2_k)^2}{\sum_{k = 1}^{n}(\sigma^2_k)^2 + \sum_{k \neq j}\sigma^2_k\sigma^2_j}$$
Having all variances strictly positive and finite makes this limit zero which sandwiches the Lindeberg condition to zero.  Where is the mistake here? 
(Hmmm... except if what is needed for the last limit to go to zero is that all variances are not just finite, but also bounded, which is not part of the premises)
 A: User @CliffAB spotted the unjustified step here, which is to assume that the conditional variance is never greater than the unconditional one.  
A quick simulation: $100,000$ draws from a standard normal, $Z$.
Sample standard deviation of the full sample : $0.997$.
Now, cut out the body of the distribution and keep the tails (which is in the spirit of the conditioning event in the Lindeberg condition) : specifically, keep only the draws where $|Z|>1.65$. Survived: $9,805$ draws. Standard deviation of them : $2.097$. 
Intuition : when we keep both tails, it is as though we tend to transform a continuous distribution into a discrete one, creating two lumps of probability mass, and this tends to increase the variance. Consider the Bernoulli $(0.5)$ distribution and the Uniform $(0,1)$. The first is the extreme discrete version of the second, and it has three times greater variance ($1/4$ as opposed to $1/12$). And the evolution is gradual : consider a discrete uniform taking values $\{0,0.25,0.75,1\}$. Its variance is $9/64$, which is inbetween the previous two.
