If $E(X_n^2) = \infty$ then $\limsup \frac{|X_n|}{\sqrt{n}} \geq a$ almost surely. 
We have given $X_1,X_2,\ldots$ an i.i.d. sequence of random variables such that $$\Bbb{E}(X_1^2)=\infty$$ I claim that for all $a>0$ $$\Bbb{P}\left(\limsup_{n\rightarrow \infty} \frac{|X_n|}{\sqrt{n}}\geq a\right)=1$$

My idea was to use Borel-Cantelli, but somehow I'm a bit confused since I never used that $\Bbb{E}(X_1^2)=\infty$.
I wanted to do this as follows:

Let $\Lambda_n=\left\{\frac{|X_n|}{\sqrt{n}}\geq a\right\}$ then $$\sum_{n=1}^\infty \Bbb{P}(\Lambda_n)=\sum_{n=1}^\infty \Bbb{P}\left(\frac{|X_n|}{\sqrt{n}}\geq a\right)=\sum_{n=1}^\infty 1-\Bbb{P}\left(\frac{|X_n|}{\sqrt{n}}\leq a\right)$$Now if $$\sum_{n=1}^\infty 1-\Bbb{P}\left(\frac{|X_n|}{\sqrt{n}}\leq a\right)<\infty$$ then it would mean that for infinitely many $n\in \Bbb{N}$ $$\Bbb{P}\left(\frac{|X_n|}{\sqrt{n}}\leq a\right)=1$$ Here I think I need some argument to show that this is not possible right?

If this works I then could apply Borel-Cantelli and would be done.
I'm not sure if this is correct so.
(I also thought about the central limit theorem but I don't think this is useful here)

Note that for a non-negative random variable $Y$, we have $$E(Y) = \int_0^\infty P(Y > y) dy$$
Since $S(y) = P(Y > y) = 1 - F_Y(y)$ is a decreasing function in $y$, we have the following Riemann sum approximation:
$$E(Y) = \int_0^\infty P(Y > y) dy \leq \sum_{n=1}^\infty P(Y \geq n)$$
Define the events $\Lambda_n$ as you did: i.e. $\Lambda_n = \{ |X_n| \geq a \sqrt{n} \} = \left\lbrace \left(\frac{X_n}{a} \right)^2 \geq  n\right\rbrace$.
Then,
$$\begin{align*}
\sum_{n=1}^\infty P(\Lambda_n) &= \sum_{n=1}^\infty P\left(\left\lbrace \left(\frac{X_n}{a} \right)^2 \geq  n\right\rbrace \right)\\
&= \sum_{n=1}^\infty P\left(\left\lbrace \left(\frac{X_1}{a} \right)^2 \geq  n\right\rbrace \right) \\
&\geq \int P(Y >y)dy & \text{where }Y = \frac{X_1^2}{a^2} \\
&= E(Y) \\
&= \infty
\end{align*}$$
Since the events $\Lambda_n$ are independent, the second Borel-Cantelli lemma allows you to conclude that $P\left( \limsup_{n \to \infty} \Lambda_n \right) = 1$.
