= Solution
The \b[<Lévy characterization of Brownian motion>] states that an <adapted> continuous real process $M$ with $M_0=0$ is standard <Brownian motion> relative to $(\mathcal F_t)$ if and only if it is a <continuous local martingale> and $[M]_t=t$. The multidimensional version replaces the bracket identity by $[M^i,M^j]_t=\delta_{ij}t$.
For the converse direction, a standard <Brownian motion> is a continuous <martingale>, and its independent centered increments give <quadratic variation> $t$. Distinct independent coordinates have zero <quadratic covariation>.
For the substantive direction, fix a vector $\theta\in\mathbb R^d$. The <Itô formula> applied to
$$
E_t(\theta)=\exp\{i\theta\cdot M_t+\tfrac12|\theta|^2t\}
$$
shows that its drift is zero: the second-order term is $-\tfrac12|\theta|^2E_tdt$ and cancels the time derivative. Thus $E(\theta)$ is a complex <local martingale>. On $[0,T]$ its modulus is the deterministic value $e^{|\theta|^2t/2}\leq e^{|\theta|^2T/2}$; localization and the <dominated convergence theorem> make it a true <martingale>. Hence
$$
\boxed{\mathbb E[e^{i\theta\cdot(M_t-M_s)}\mid\mathcal F_s]=e^{-\frac12|\theta|^2(t-s)}.}
$$
This conditional <characteristic function> is the deterministic <characteristic function> of $N(0,(t-s)I)$. For $A\in\mathcal F_s$, multiplying by $\mathbf1_A$ and taking <expectations>, then using uniqueness of <characteristic functions>, proves that $M_t-M_s$ has this <multivariate normal distribution> and is independent of $\mathcal F_s$. Iterated conditioning gives independent increments with <Gaussian distributions>. Continuity, the zero initial value and these increment laws are the definition of standard <Brownian motion>. The case $d=1$ proves the requested statement; the vector form will also be used below.
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