Using the fact that $S:=C[0,+\infty):=(f\colon [0,+\infty)\to \Bbb R, f\mbox{ continuous })$ endowed with the metric $$d(f,g):=\sum_{j=1}^{+\infty}2^{—j}\min(1,\sup_{0\leq t\leq j}|f(x)-g(x)|)$$ is polish, it's enough to show that each subsequence is tight. Using theorem 4.10 in Karatzas-Shreve book's, we know that a sequence of probability measures $(\mathbb P_n)_{n\geq 1}$ over the Borel $\sigma$-albegra is tight if and only if the two following conditions are satisfied:

- $\lim_{\lambda\to+\infty}\sup_{n\geq 1}\mathbb P_n(\omega,|\omega(0)|\geq \lambda)=0$ and
- $\lim_{\delta\to 0}\sup_{n\geq 1}\mathbb P_n(\omega,m^T(\omega,\delta)\geq \varepsilon)=0$ for each $T>0$ and $\varepsilon>0$, where $m^T(\omega,\delta)=\sup_{|t_1-t_2|\leq\delta,0\leq t_1,t_2\leq T}|\omega(t_1)-\omega(t_2)|$.

We apply this to $\mathbb P_n=\mathbb P_{x_n}$, where $(x_n)_{n\geq 1}$ is a sequence in $D$. By boundedness of $D$, the first condition is satisfied. We can control the modulus of continuity using the fact that $$\mathbb P_n(\omega,m^T(\omega,\delta)\geq \varepsilon)
\leq \frac 1{\varepsilon^2}\delta^2,$$
which is a consequence of the fact that the increments of the Brownian motion $W_{t_2}-W_{t_1}$ are normally distributed, of normal distribution of mean $0$ and variance $t_2-t_1$.

In fact, with condition 1., we can see that $(\mathcal P_x,x\in D)$ is tight (where $\mathcal P_x$ is associated to a Brownian motion started at $x$) if and only if $D$ is bounded.