If $E|X_i|^{2}\rightarrow0$, $\frac{S_n}{n}\xrightarrow{p}0$ is not always true.

We know $E|X_i|^{2}\rightarrow0 \implies X_n\xrightarrow{p}0$. Proof: using Chebyshev's inequality, $P(|X_n|>\epsilon)<\epsilon^{-2}E|X_n|^2\rightarrow0$.

Let $S_n = \sum_{m=1}^{n}X_i$. Since we do not have restrictions on $X_1, X_2, ...$, I can construct a counterexample by letting $X_1=c, c\in\mathbb{R}, c>0$. Obviously, $\frac{S_n}{n}$ would not converge to $0$ in probability at this point. Is this a correct idea?

It is true, because $$E\,\left\lvert\frac{X_1+\cdots+X_n}n\right\rvert^2\le\frac1n\sum_{i=1}^nE\,|X_i|^2\xrightarrow[n\to\infty]{}0$$ by Jensen's inequality and Cesàro-Stolz theorem, and because convergence in $L^2$ entails convergence in probability.