CLT cannot be enhanced to convergence in probability 
Let $\{ \xi_n \}^{\infty}_{n=1}$ be iid nondegenerate random variables with finite second moment. Let $\mathbb{E} \xi_i = a$, $S_n = \xi_1 + \cdots + \xi_n$. Prove that $\sqrt{n} \left( \dfrac{S_n}{n} - a\right)$
  converges in distribution and has no limit for the convergence in probability.

Сonvergence in distribution is proved, using the central limit theorem. How to prove that there is no convergence in probability?
 A: There may be a more elementary solution, but one way to show this is to note that convergence in probability implies convergence almost surely along a subsequence. Define $A_n:=\sqrt n \big(\frac{S_n}{n} -a\big)$. If $A_n \to A$ is probability, then there is a subsequence $A_{n_k} \to A$ almost surely.
There are now two methods which can be used to derive a contradiction here:
Method 1: On one hand, $A$ must be normally distributed since convergence in probability implies convergence in distribution. On the other hand, it is easily shown that $A$ is measurable with respect to the tail $\sigma$-algebra of the increments of $S_n$. This is indeed the case because for any fixed $N$, clearly $\big(\frac{S_n }{\sqrt n} - a \sqrt n\big)$ and $\big(\frac{S_n-S_N}{\sqrt n}-a\sqrt n\big)$ will have the same subsequential limit as $n \to \infty$. Consequently $A$ must be constant, by Kolmogorov's $0$-$1$ law. This contradits the fact that $A$ was normally distributed with positive variance.
Method 2: To obtain a contradiction, we will show that the random set $\{A_{n_k}: k \in \Bbb N\}$ must be dense in $\Bbb R$. To prove this, we mimic the argument given in the answers here: Normalized partial sums of normal random variables are dense in $\mathbb{R}$
Specifically, for real numbers $a<b$ we let $E_{a,b}$ denote the event that $A_{n_k} \in [a,b]$ infinitely often. It is easily checked that $E_{a,b}$ is a tail event (and also an exchangeable event) for the increments of $S_n$, and consequently $P(E_{a,b}) \in \{0,1\}$ (by Kolmogorov's or Hewitt-Savage's $0$-$1$ law). But by (reverse) Fatou's Lemma we also have that $$P(E_{a,b}) = P\big( \limsup_k \{A_{n_k} \in [a,b]\}\big) \geq \limsup_k P(A_{n_k}\in [a,b]) =P(Z\in[a,b])>0$$ where $Z$ is normally distributed and the final equality follows from the central limit theorem. So we see that $P(E_{a,b})=1$. Consequently we get that $P(\{A_{n_k}\}$ is dense$)=P\big( \bigcap_{a<b \in \Bbb Q} E_{a,b}\big)=1$. So we have a contradiction, since convergent sequences cannot have dense image in $\Bbb R$.
A: A possible way is the following. 

Assume that $\left(Z_n\right)$ and $(Z'_n)$ are two sequences of random variables such that
  
  
*
  
*For any $n\geqslant 1$, $Z_n$ is independent of $Z'_n$. 
  
*The sequence $\left(Z_n+Z'_n\right)_{n\geqslant 1}$ converges to $0$ in probability. 
  
*For any $R>0$, there exists $ \delta\gt 0$  and $n_0(R)$ such that for any $n\geqslant n_0(R)$, $\mathbb P\left(\left|Z'_n\right|\leqslant R\right)\gt \delta$.
  
  
  Then the sequence $(Z_n)$ converges to $0$ in probability.

To see this, let $\varepsilon\gt 0$. Then the following inclusion holds:
$$\left\{\left|Z_n+Z'_n\right|\geqslant \varepsilon\right\}\supset\left\{\left|Z_n\right|\geqslant 2\varepsilon\right\}\cap \left\{\left| Z'_n\right|\leqslant \varepsilon\right\}.$$
Taking the probabilities on both sides and using independence, we derive that for $n\geqslant n_0(\varepsilon)$, $$\mathbb P\left\{\left|Z_n+Z'_n\right|\geqslant \varepsilon\right\}\geqslant\mathbb P\left\{\left|Z_n\right|\geqslant 2\varepsilon\right\}\delta.$$
To conclude, use this with $$Z_n=\frac{S_{2n}-S_n  }{2\sqrt n}\mbox{ and }Z'_n:=\left(\frac 1{\sqrt 2}-1\right)\frac{S_n}{\sqrt n}.          $$
