please confirm if the following is correct.
Let $A = \{V_\tau > \epsilon\}$ and $\alpha = \min(\alpha_1,\tau)$, where $\alpha_1 = \min\{t\geq 0: X_t\geq \epsilon\}$ (So intuitively, $\alpha$ is a stopping time since it is the minimum of stopping times). Now consider the following inequalities:
$$\begin{align} E(A_\tau)\geq E(A_\alpha)\quad &\textrm{(since $A$ is increasing and $\alpha\leq \tau)$} \\ E(A_\alpha)\geq E(X_\alpha) \quad &\textrm{(given, since $\alpha$ is a stopping time)}.\end{align} $$ So we have $$\begin{align}E(A_\tau)\geq E(A_\alpha)\geq E(X_\alpha)&=\int_{\Omega} X_\alpha \,\mathrm dP=\int_{A} X_\alpha \,\mathrm dP + \int_{A^\prime} X_\alpha \,\mathrm dP\\ &\geq \int_{A} \epsilon \,\mathrm dP + \int_{A^\prime} X_\alpha \,\mathrm dP \\&\geq \epsilon \Pr(A)\end{align} $$ and hence the result.