I am having trouble to understand the following. If $S_n=X_1+X_2+......+X_n$, where X_1,X_2 are Bernouli (p). 
I don't understand this. So you get an intermediate point Constant* sqrt(n). To the right of this intermediate point, you get Probability =1. To the left of this, you get Probability=0. Is that how you interpret this? (refer to the picture below).
This is an asymptotic result assuming independence of trials.
Use CLT to show that $\sqrt n((\frac{1}{n}\sum X_i)-p)=O_p(1)$. So, $\bar S_n=\frac{S_n}{n}$ is $\sqrt n$ bounded. Hence, if you change the rate of convergence slightly, it becomes unbounded (your limit goes to 1) or converges in probability to 0 (in this case you get probability 0 in the limit).
By CLT I mean that for any $c>0$
$\lim_{n \to \infty}P\{|S_n-np|\le \sqrt nc\}=\lim_{n\rightarrow \infty}P\{|\frac {1}{\sqrt n}\sum (X_i-p)|\le c\}=F(c)-F(-c)=Z(c)$
where $F$ is CDF of normal distribution with zero mean and appropriate variance. Then for any $\epsilon>0$
$\lim_{n\rightarrow \infty}P\{|S_n-np|\le n^{\frac{1}{2}+\epsilon}c\}=\lim_{n\rightarrow \infty}P\{|\frac {1}{\sqrt n}\sum (X_i-p)|\le n^\epsilon c\}=Z(\infty)=1$
$\lim_{n\rightarrow \infty}P\{|S_n-np|\le n^{\frac{1}{2}-\epsilon}c\}=\lim_{n\rightarrow \infty}P\{|\frac {1}{\sqrt n}\sum (X_i-p)|\le \frac{c}{n^\epsilon}\}=Z(0)=0$
Another interpretation is that the deviation of a sum of Bernouli trials from its mean accumulates with a specific rate ($\sqrt n$). As $n\rightarrow \infty$ all accumulated sums fall inside $\pm \sqrt n$ bounds (solid black lines in the graph below). Compare this with slightly different bounds (dotted lines): $\pm n^{\frac{1}{2}\pm 0.01}$