= Solution
Use the independent-mark interpretation: the <Brownian motions> attached to the atoms are <independent> of one another and of the initial <Poisson random measure>. Fix $t$ and define
$$
p_t(x)=\mathbb P\bigl(\exists s\in[0,t]:|x+\xi(s)-f(s)|\leq1\bigr).
$$
Conditionally on the initial atoms, each atom is retained as a dangerous atom if its own path meets the target by time $t$. These tests are <independent>, with position-dependent retention <probability> $p_t(x)$. The permitted <Poisson thinning> property makes the dangerous atoms a <Poisson random measure> of intensity $p_t(x)\,dx$.
Its total intensity is computed without invoking a marking or displacement theorem. The event in the definition of $p_t(x)$ says that $x$ belongs to
$$
W_-(t)=\bigcup_{0\leq s\leq t}\mathcal B(f(s)-\xi(s),1).
$$
By the <Tonelli theorem>,
$$
\int_{\mathbb R^d}p_t(x)\,dx=\mathbb E\int_{\mathbb R^d}\mathbf1_{\{x\in W_-(t)\}}\,dx=\mathbb E\operatorname{vol}(W_-(t)).
$$
Symmetry of <Brownian motion> gives $-\xi\stackrel d=\xi$ as processes. It therefore identifies the last <expectation> with $\mathbb E\operatorname{vol}(W(t))$, where the sign of the deterministic path $f$ is unchanged. This swept set is a <Wiener sausage> with an added deterministic path.
The total intensity is finite on each bounded time interval. To see this explicitly, write $K_t=1+\sup_{s\leq t}|f(s)|$ and $R_t=\sup_{s\leq t}|\xi(s)|$. The sausage is contained in a ball of radius $K_t+R_t$. On a finite time grid, splitting according to the first crossing of a positive level $u$ by a one-dimensional <Brownian motion> shows that the chance of ending above $u$, conditional on the crossing, is at least $1/2$: the remaining <independent> increment is symmetric. Thus the grid maximum exceeds $u$ with <probability> at most twice the final Gaussian upper tail. Dense grids and continuity give the same bound for the path maximum. Applying this to each coordinate and its negative gives
$$
\mathbb P(R_t>u)\leq4d\,\mathbb P(N(0,t)>u/\sqrt d)\leq4d\,e^{-u^2/(2dt)}\qquad(t>0).
$$
The last inequality follows from $\mathbb E e^{\theta N(0,t)}=e^{\theta^2t/2}$ and minimizing the exponential Markov bound. Integration of this tail gives finite moments of $R_t$ of every positive order, so $\mathbb E\operatorname{vol}(W(t))<\infty$.
There are consequently only finitely many dangerous atoms on each compact time interval, almost surely. For closed balls, continuity of the paths makes each finite hitting-time infimum attained. The event $T>t$ is therefore the absence of dangerous atoms up to time $t$.
The open-ball convention has the same survival <probability> at each fixed $t$. Indeed for any compact path range $K$, the difference between its closed and open radius-one neighborhoods is the shell $\{x:\operatorname{dist}(x,K)=1\}$, which has zero <Lebesgue measure>. To see this, choose a nearest $y\in K$ for a shell point $x$. For every sufficiently small $\varepsilon>0$, the ball of radius $\varepsilon/2$ centered at $x-\varepsilon(x-y)$ lies in the open neighborhood and inside the ball of radius $3\varepsilon/2$ about $x$. Thus the shell has a fixed proportional hole at every scale; the <Lebesgue density theorem> forces its measure to be zero. By the <Tonelli theorem>, closed-touch and open-entry retention probabilities have equal integrals. Independent thinning therefore gives no atom that touches only the boundary before this fixed horizon, almost surely. This also excludes a difference at an open-entry infimum equal to $t$.
The zero-count <probability> of a <Poisson distribution> is $e^{-\lambda}$, and hence
$$
\boxed{\mathbb P(T>t)=\exp\{-\mathbb E\operatorname{vol}(W(t))\}.}
$$
This <survival among independently moving Poisson traps> formula has been derived using only <independent> thinning and the defining Poisson zero-count <probability>, with the <expectation>/volume step supplied by Tonelli's theorem.
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