There are two parts to this problem:
1) $\limsup_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}] \leq E[|Z|] $
Proof: (Credits to DavideGiraudo)
$E[\frac{|S_n|}{\sqrt{n}}] = E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} \leq K}] + E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} > K}]$ $\leq E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} \leq K}] + \sqrt{E[\frac{S_n^2}{n}]}.\sqrt{P(\frac{|S_n|}{\sqrt{n}} > K) } $ (Cauchy Schwarz) $\leq E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} \leq K}] + \sigma^2.\frac{1}{K} $ (Used the Markov Inequality)
Take limsup on both sides
$\limsup_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}] \leq \limsup_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} \leq K}] + \sigma^2.\frac{1}{K}$ $ \leq \limsup_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} \leq K}] + K.P(\frac{|S_n|}{\sqrt{n}} > K) + \sigma^2.\frac{1}{K} \quad (\times) $ I have added that extra term for a reason...
Now use the following: (a) $f(x)=\min(|x|,K)$ is continuous and bounded
(b) $E[f(X_n)] \rightarrow E[f(X)] $ if $X_n \rightarrow^w X$. f as in (a)
Note that $ E[f(\frac{|S_n|}{\sqrt{n}})] = E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} \leq K}] + K.P(\frac{|S_n|}{\sqrt{n}} > K)$
Now apply (a) and (b) to $(\times)$ to get
$\limsup_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}] \leq E[|Z|1_{|Z| \leq K}] + K.P(|Z| > K) + \sigma^2.\frac{1}{K}$ LHS is independent of K. Take $K \rightarrow \infty$. Note that $K.P(|Z| > K) \rightarrow 0$ and MCT applied to the first term yields
$\limsup_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}] \leq E[|Z|] $
(2) $\liminf_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}] \geq E[|Z|] $ Proof:
$ E[\frac{|S_n|}{\sqrt{n}}] = E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} \leq K}] + E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} > K}] $ $ \geq E[\frac{|S_n|}{\sqrt{n}}1_{\frac{|S_n|}{\sqrt{n}} \leq K}] + K.P(\frac{|S_n|}{\sqrt{n}} > K)$ Take $\liminf_{n \rightarrow \infty}$ to get
$\liminf_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}] \geq E[|Z|1_{|Z| \leq K}] + K.P(|Z| > K)$ (I used (b)) Take $K \rightarrow \infty$, use MCT on the first term on RHS to get
$\liminf_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}] \geq E[|Z|] $
Thus we get $\lim_{n \rightarrow \infty} E[\frac{|S_n|}{\sqrt{n}}] = E[|Z|] $ QED