= Solution
Take
$$
\boxed{\mathcal D(\Delta/2)=C_c^\infty(\mathbb R^d).}
$$
Part (i) proves that <Brownian motion> solves this <martingale problem>. For the converse, let $X$ be any continuous <adapted process> satisfying this <martingale problem> and $X_0=0$.
Set $\tau_n=\inf\{t\geq0:|X_t|\geq n\}$. Choose a <smooth cutoff function> $\chi_n$ equal to one on a neighbourhood of the closed ball of radius $n$. Apply the <martingale problem> to $\chi_n(x)x_i$ and $\chi_n(x)x_ix_j$, and stop at $\tau_n$. Inside that ball, their <Laplace operators> are respectively $0$ and $2\delta_{ij}$. The <optional sampling theorem> therefore shows that
$$
X^i_{t\wedge\tau_n},\qquad
X^i_{t\wedge\tau_n}X^j_{t\wedge\tau_n}-\delta_{ij}(t\wedge\tau_n)
$$
are <martingales>. Continuity makes $\tau_n\uparrow\infty$ <almost surely>. Hence every $X^i$ is a <continuous local martingale>. The <Itô product rule> and the fact that a <continuous finite-variation local martingale is constant> identify its <quadratic covariations>:
$$
[X^i,X^j]_t=\delta_{ij}t.
$$
Here $\delta_{ij}$ is the <Kronecker delta>.
The <Lévy characterization of multidimensional Brownian motion> now applies. To see explicitly why it identifies the entire law, for $\xi\in\mathbb R^d$ the <Itô formula> makes
$$
E_t(\xi)=\exp\!\left(i\xi\cdot X_t+\frac12|\xi|^2t\right)
$$
a complex <local martingale>. Its absolute value on $[0,T]$ is at most $e^{|\xi|^2T/2}$, so the <bounded local martingale criterion> makes it a true <martingale> there. For $0\leq s\leq t$,
$$
\boxed{\mathbb E\!\left[e^{i\xi\cdot(X_t-X_s)}\mid\mathcal F_s\right]
=e^{-|\xi|^2(t-s)/2}.}
$$
This conditional <characteristic function> identifies $X_t-X_s$ as having a centered <multivariate normal distribution> with <covariance matrix> $(t-s)I$, <independent> of $\mathcal F_s$. Together with continuity and $X_0=0$, these are exactly the defining properties of standard $d$-dimensional <Brownian motion>. Thus the <martingale problem for Brownian motion> has a unique law.
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