Let $t_0\in\mathbb R$, $f:[t_0,\infty)\to\mathbb R$ be continuous, $B\subseteq\mathbb R$, $I:=\{t\ge t_0:f(t)\in B\}$, $\tau:=\inf I$ and $$g(t):=f(\min(t,\tau))\;\;\;\text{for }t\ge t_0.$$
If$^1$ $f(t_0)\le\inf B$, are we able to show that $g(t)\not\in B$ for all $t\ge t_0$?
If $I=\emptyset$, then $\tau=\infty$ and $g(t)=f(t)\not\in B$ for all $t\ge t_0$. So, assume $I\ne\emptyset$ and let $t_B\in I$. Then $\tau\in[t_0,\tau_B]$ and $f(t_0)\le\inf B\le f(t_B)$. Since $f$ is continuous, the intermediate value theorem yields that there is a $t_{\inf B}\in[t_0,t_B]$ with $$f(t_{\inf B})=\inf B\tag1.$$
At this point, we might need an additional assumption. I would like to assume something like $$f(\tau)>\inf B\tag2$$ and show that this yields a contradiction, but I'm not sure if this is possible in general.
So, which additional assumption do we need?
If, for example, $B=(a,\infty)$ for some $a\in\mathbb R$, then $(2)$ would yield (by another application of the intermediate value theorem) that there is a $t\in(t_B,\tau)$ with $X_t>a=\inf B$ and hence $t\in I$; in contradiction to the definition of $\tau$.
$^1$ Please note that the only reason I'm imposing this assumption is that it justifies the first application of the intermediate value theorem in my proof attempt.
There are trivial counter-examples. Take $t_0=0, f(x)=x$ and $B=\{0\}$.