= Solution
Write $e_t=\mathbf1_{[0,t]}$ as an element of <L2 space>. Parts (iii) and (iv) show that $B_t=We_t$ is a centered <Gaussian process> with <covariance function>
$$
\mathbb E[B_sB_t]=\langle e_s,e_t\rangle=\min(s,t).
$$
In particular $B_t-B_s\sim N(0,|t-s|)$. The fourth <moment> of a centered <normal random variable> of <variance> $v$ is $3v^2$, so
$$
\mathbb E|B_t-B_s|^4=3|t-s|^2.
$$
The <Kolmogorov continuity theorem>, with moment exponent $4$ and time exponent $2>1$, provides a <continuous modification> $\widehat B$. At each finite set of times this <modification of a stochastic process> has the same joint law as $B$.
For $0\leq t_0<t_1<\cdots<t_m\leq1$, the successive increments are <Jointly Gaussian>. The <covariance> of two increments on disjoint intervals equals the <inner product> of their disjoint interval indicators and is zero. Zero cross-<covariances> of <Jointly Gaussian> variables imply <independence>, and each increment has <variance> $t_j-t_{j-1}$. Finally $e_0=0$ in <L2 space>, so $\widehat B_0=0$ <almost surely>. Therefore
$$
\boxed{\widehat B\text{ is standard Brownian motion on }[0,1].}
$$
Independence from the whole past in its <natural Brownian filtration> follows by extending finite-dimensional <independence> to the generated <sigma-algebra>.
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