No, it is not true that a process *W* satisfying the properties (1), (3) and (4) has to be a Brownian motion. We can construct a counter-example as follows.

This construction is rather contrived, and I don't know if there's any simple examples.
Start with a standard Brownian motion *W*. The idea is to apply a small bump to its distribution while retaining the required properties. I will do this by first reducing it to the discrete-time case. So, choose a finite sequence of times 0 = *t*_{0} < *t*_{1} < ... < *t*_{n}. Then define a piecewise linear process *X* by *X*_{tk} = *W*_{tk} (*k* = 0,1,...,*n*) and such that *X* is linearly interpolated across each of the intervals [*t*_{k-1},*t*_{k}] and constant over [*t*_{n},∞).

Then, *Y* = *W* - *X* is a continuous process independent from *X*. In fact, *Y* is just a sequence of Brownian bridges across the intervals [*t*_{k-1},*t*_{k}] and is a standard Brownian motion on [*t*_{n},∞). Also by linear interpolation, for any time *t* ≥ 0, *X*_{t} is a linear combination of at most two of the random variables *X*_{t1},...,*X*_{tn}. The increments of *W*,
$$
W_t-W_s = X_t-X_s + Y_t-Y_s,
$$
are then a linear combination of at most 4 of the random variables *X*_{t1},...,*X*_{tn} plus an independent term. So, choosing *n* ≥ 5, if it is possible to replace (*X*_{t1},...,*X*_{tn}) by any other ℝ^{n}-valued random variable without changing the joint-distribution of any 4 elements, then the distributions of the increments *W*_{t} - *W*_{s} will be left unchanged. So, properties (1), (3), (4) will still be satisfied but the new process for *W* will not be a standard Brownian motion. It is possible to change the distribution in this way:

Let *X* = (*X*_{1},*X*_{2},...,*X*_{n}) be an ℝ^{n}-valued random variable with a continuous and strictly positive probability density *p*_{X}: ℝ^{n} → ℝ. Then, there exists a random variable *Y* = (*Y*_{1},*Y*_{2},...,*Y*_{n}) with a different distribution than *X* but for which the projection onto any *n* - 1 elements has the same distribution as for *X*.

That is, for any *k*_{1},*k*_{2},...,*k*_{n-1} in {1,...,*n*}, (*Y*_{k1},*Y*_{k2},...,*Y*_{kn-1}) has the same distribution as (*X*_{k1},*X*_{k2},...,*X*_{kn-1}).

We can construct the probability density *p*_{Y} of *Y* by applying a bump to the probability distribution of *X*,
$$
p_Y(x)=p_X(x)+\epsilon f(x_1)f(x_2)\cdots f(x_n).
$$
Here, ε is a fixed real number and *f*: ℝ → ℝ is a continuous function of compact support and zero integral, $\int_{-\infty}^\infty f(x)\\,dx=0$. Then, $\int_{-\infty}^\infty p_Y(x)\\,dx_k=\int_{-\infty}^\infty p_X(x)\\,dx_k$ for each *k*. So, the integral of *p*_{Y} over ℝ^{n} is 1 and, by choosing ε small, *p*_{Y} will be positive. Then it is a valid probability density function. Finally, as the integral along the *k*^{th} direction (any *k*) agrees for *p*_{X} and *p*_{Y}, the projection of *X* and *Y* onto ℝ^{n-1} along the *k*^{th} direction give the same distribution.