Setup
Assume the following formation $$ \bar{X} \overset{P}{\to} 0 \text{ as } n\to\infty. $$
Lemma
We want to show $$ \sqrt{n}\cdot\left(\bar{X}\right)^2 \overset{P}{\to} 0. $$ This is indeterminate form ($\infty \times 0$).
It seems to me that the sample mean would win because it is sensibly multiplied twice by something that converges to zero, but how can this be shown mathematically?
Background
Let $X_1, \ldots, X_n$ be i.i.d. $N(\xi, \sigma^2)$ and consider the problem of simultaneously estimating or testing $\xi$ and $\sigma^2$. Then, we are interested in the joint distribution of $$ Y_1 := \frac{\sqrt{n}(\bar{X}-\xi)}{\sigma}\text{ and }Y_2 := \sqrt{n} \left( \frac{\sum_{i=1}^n (X_i-\bar{X})^2}{n\sigma^2}-1\right). $$ Since this distribution is independent of $\xi$ and $\sigma$, suppose that $\xi = 0$. By bivariate CLT, $$ \frac{\sqrt{n}\bar{X}}{\sigma}\text{ and }\sqrt{n} \left( \frac{\sum_{i=1}^n X_i^2}{n\sigma^2}-1\right) $$ are asymptotically independently distributed according to $N(0, 1)$ and $N(0, 2)$, respectively. At this point, if the Lemma holds, we can get $$ \begin{pmatrix} Y_1 \\ Y_2 \end{pmatrix}\overset{d}{\to} N_2\left( \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 & 0\\ 0 & 2 \end{pmatrix} \right) $$ from Slutsky's theorem.
Here I assume the following:$(X_1, \ldots, X_n)$ is a random iid sample with $\mathbb{E}(X_i) = 0, \text{Var}(X_i) = \sigma^2 < \infty$.
For $\epsilon > 0$, by the Markov's inequality, we have $$ \mathbb{P}(\sqrt{n}(\bar{X})^2 > \epsilon) \le \dfrac{\sqrt{n}}{\epsilon}\mathbb{E}[(\bar{X})^2] $$ But, $$ \mathbb{E}[(\bar{X})^2] = \text{Var}(\bar{X}) = \dfrac{\sigma^2}{n} $$ Thus, $$ \mathbb{P}(\sqrt{n}(\bar{X})^2 > \epsilon) \le \dfrac{\sigma^2}{\sqrt{n}\epsilon} \rightarrow 0, \text{ as } n \rightarrow \infty $$