(Post edited after the suggestion from the first comment.) The following simple result is, I suppose, probably standard, even though I did not find a specific reference neither a previous thread here in the forum. What I ask is for a proof-check, that is wether or not I am doing something wrong (or a reference in which is stated correctly and proved).
Suppose $X_t\left(m\right)$ is a sequence of stochastic processes and $\tau_n$ a sequence of stopping times such that $\mathbb{P}\left[\tau_n\geq t\right]\rightarrow 1$ as $n\rightarrow+\infty$ (for any $t\geq 0$). Suppose that, for any $t\geq 0$ and for any $n$, it holds that
$$
\lim_{m\rightarrow+\infty}\mathbb{E}\left[X_{t\wedge\tau_n}\left(m\right)^2\right] = 0,
$$
i.e. $X_t^{(\tau_n)}(m)\stackrel{L^2}{\longrightarrow}0$ as $m\rightarrow +\infty$, where $X_t^{(\tau_n)}(m)$ is the stopped versions of $X_t(m)$. I want to find under which conditions
$$
X_t(m)\stackrel{p}{\longrightarrow}0,\quad\text{ as }m\rightarrow+\infty,
$$
that is the convergence holds in probability for the unstopped process.
To do this I try to split \begin{eqnarray} \mathbb{P}\left[\left|X_t(m)\right|\geq \varepsilon\right]&=&\mathbb{P}\left[\{\left|X_t(m)\right|\geq \varepsilon\}\cap\{\tau_n<t\}\right]+\mathbb{P}\left[\{\left|X_t(m)\right|\geq \varepsilon\}\cap\{\tau_n\geq t\} \right] \\ &=&\mathbb{P}\left[\{\left|X_t(m)\right|\geq \varepsilon\}\cap\{\tau_n<t\}\right]+\mathbb{P}\left[\{\left|X_{t\wedge\tau_n}(m)\right|\geq \varepsilon\}\cap\{\tau_n\geq t\} \right]\\ &\leq& \mathbb{P}\left[\tau_n<t\right]+\mathbb{P}\left[\left|X_{t\wedge\tau_n}(m)\right|\geq \varepsilon \right]\quad (1). \end{eqnarray} Now $X_t^{(\tau_n)}(m)\stackrel{L^2}{\longrightarrow}0$ as $m\rightarrow+\infty$ implies $X_t^{(\tau_n)}(m)\stackrel{p}{\longrightarrow}0$ for any given $n$, which means $\mathbb{P}\left[\left|X_{t\wedge\tau_n}(m)\right|\geq \varepsilon \right]\rightarrow 0$ as $m\rightarrow+\infty$. Hence if, for given $n$, I take the limit $m\rightarrow+\infty$ on both side of $(1)$ I get $$ \overline{\lim}_{m\rightarrow+\infty}\mathbb{P}\left[\left|X_t(m)\right|\geq \varepsilon\right] \leq \mathbb{P}\left[\tau_n<t\right]\quad (2). $$ To conclude I take the limit ${n\rightarrow+\infty}$ on both sides of $(2)$. Since the left hand side is constant and the right hand side goes to zero I find $$ \overline{\lim}_{m\rightarrow+\infty}\mathbb{P}\left[\left|X_t(m)\right|\geq \varepsilon\right] =0, $$ Nevertheless the sequence $s_{m}\doteq\mathbb{P}\left[\left|X_t(m)\right|\geq\varepsilon\right]$ is such that $0\leq s_m\leq 1$, whence $$ 0\leq \underline{\lim}_{m\rightarrow+\infty}s_m\leq \overline{\lim}_{m\rightarrow+\infty}s_m = 0 $$ so $$ \lim_{m\rightarrow+\infty}\mathbb{P}\left[\left|X_t(m)\right|\geq\varepsilon\right]=0. $$