Let $(\tau_n)_{n\in\mathbb N}\subseteq\overline{\mathbb R}$, $$\tau:=\inf_{n\in\mathbb N}\tau_n$$ and $t\in\overline{\mathbb R}$. Are we able to show that $$\tau<t\Leftrightarrow\exists n\in\mathbb N:\tau_n<t\tag1$$ and if so, does $(1)$ remain true when "$<$" is replaced by "$\le$"?
Of course, "$\Leftarrow$" is trivial; no matter we consider "$<$" or "$\le$".
But I struggle to obtain "$\Rightarrow$". Clearly, by definition of the infimum, for all $\varepsilon>0$ there is a $n_0\in\mathbb N$ with $$\tau_{n_0}<\tau+\varepsilon\tag2,$$ but since $n_0$ depends on $\varepsilon$, I'm not sure how to conclude.
If $\tau<t$, we can simply let $\varepsilon:=(t-\tau)/2>0$. As you have mentioned in your post, there is $n\in\mathbb{N}$ with $$\tau_n<\tau+\varepsilon=\tau+\frac{t-\tau}{2}=\frac{t+\tau}{2}<t.$$
However, if you assume that $\tau\leq t$, this implication may fail as the possibility for $\tau=t$ exists. If $\tau=t$, then apparently, there is no such $n\in\mathbb{N}$ such that $\tau_n<t=\tau$ by the definition of infimum.