For this proof we need $E|X|<\infty$ (it seems that this is the case).
Notice that
$$E(Y_k^2)\leq \int_0^k 2tP(|X|>t)dt. $$
Hence, by Fubini's theorem
\begin{align}
\sum^\infty_{k=1}\frac{E(Y_k^2)}{k^2} &\leq \int_0^\infty \left(\sum^\infty_{k=1}\frac{1_{(t\leq k)}}{k^2}\right)2tP(|X|>t)dt\\
& \int_0^\infty\left( \sum^\infty_{k=t}\frac{1}{k^2}\right)2tP(|X|>t)dt\\
& \leq \int_0^\infty \left( \sum^\infty_{k=t}\frac{2}{k(k+1)}\right)2tP(|X|>t)dt\\
& = \int_0^\infty \left( \sum^\infty_{k=t}\frac{2}{k} - \frac{2}{k+1}\right)2tP(|X|>t)dt\\
& \leq \int_0^\infty \frac{2}{t}2tP(|X|>t)dt = 4E|X|<\infty
\end{align}
So, we proved that
$$\sum^\infty_{k=1}\frac{E(Y_k^2)}{k^2} $$
converges. Then, by Kronecker's lemma,
$$\lim_{n\rightarrow \infty}\frac{1}{n^2}\sum^n_{k=1}E(Y_k^2) =0.$$
Additional thoughts: After reading the comments, I think this may help you.
You can use Kolmogorov's maximal inequality (or a Martingale in $L_2$ convergence Theorem) to prove that if
$$ \lim_{n\rightarrow \infty}\sum^n_{k=1}\frac{\text{Var}(Y_k^2)}{k^2}$$
converges, then
$$\lim_{n\rightarrow \infty}\sum^n_{k=1}\frac{Y_k -EY_k}{k} $$
converges a.s..
Hence, by Kronecker's lemma,
$$\lim_{n\rightarrow \infty}\frac{1}{n}\sum^n_{k=1}(Y_k -EY_k)=0. $$
Using Dominated Convergence Theorem, it's easy to show
$$E Y_k \rightarrow EX. $$
Therefore, by Cesàro's Lemma,
$$\frac{1}{n}\sum_{k=1}^n Y_k = EX,\ a.s. .$$
Now, you can use your Borel-Cantelli Argument to prove that the above expression implies
$$\lim_{n\rightarrow \infty}\frac{1}{n}\sum^n_{k=1}X_k=EX,\ a.s.. $$
This is basically the proof for the Strong Law of Large Numbers in the case i.i.d. with $E|X|<\infty.$