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*<sub>0</sub> < *t*<sub>1</sub> < ... < *t*<sub>*n*</sub>. Then define a piecewise linear process *X* by *X*<sub>*t*<sub>*k*</sub></sub> = *W*<sub>*t*<sub>*k*</sub></sub> (*k* = 0,1,...,*n*) and such that *X* is linearly interpolated across each of the intervals [*t*<sub>*k*-1</sub>,*t*<sub>*k*</sub>] and constant over [*t*<sub>*n*</sub>,∞).
Then, *Y* = *W* - *X* is a continuous process independent from *X*. In fact, *Y* is just a sequence of [Brownian bridges][1] across the intervals [*t*<sub>*k*-1</sub>,*t*<sub>*k*</sub>] and is a standard Brownian motion on [*t*<sub>*n*</sub>,∞). Also by linear interpolation, for any time *t* ≥ 0, *X*<sub>*t*</sub> is a linear combination of at most two of the random variables *X*<sub>*t*<sub>1</sub></sub>,...,*X*<sub>*t*<sub>*n*</sub></sub>. 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*<sub>*t*<sub>1</sub></sub>,...,*X*<sub>*t*<sub>*n*</sub></sub> plus an independent term. So, choosing *n* ≥ 5, if it is possible to replace (*X*<sub>*t*<sub>1</sub></sub>,...,*X*<sub>*t*<sub>*n*</sub></sub>) by any other ℝ<sup>n</sup>-valued random variable without changing the joint-distribution of any 4 elements, then the distributions of the increments *W*<sub>*t*</sub> - *W*<sub>*s*</sub> 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*<sub>1</sub>,*X*<sub>2</sub>,...,*X*<sub>*n*</sub>) be an ℝ<sup>n</sup>-valued random variable with a continuous and strictly positive probability density *p*<sub>*X*</sub>: ℝ<sup>n</sup> → ℝ. Then, there exists a random variable *Y* = (*Y*<sub>1</sub>,*Y*<sub>2</sub>,...,*Y*<sub>*n*</sub>) 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*<sub>1</sub>,*k*<sub>2</sub>,...,*k*<sub>*n*-1</sub> in {1,...,*n*}, (*Y*<sub>*k*<sub>1</sub></sub>,*Y*<sub>*k*<sub>2</sub></sub>,...,*Y*<sub>*k*<sub>*n*-1</sub></sub>) has the same distribution as (*X*<sub>*k*<sub>1</sub></sub>,*X*<sub>*k*<sub>2</sub></sub>,...,*X*<sub>*k*<sub>*n*-1</sub></sub>).
We can construct the probability density *p*<sub>*Y*</sub> 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*<sub>*Y*</sub> over ℝ<sup>*n*</sup> is 1 and, by choosing ε small, *p*<sub>*Y*</sub> will be positive. Then it is a valid probability density function. Finally, as the integral along the *k*<sup>th</sup> direction (any *k*) agrees for *p*<sub>*X*</sub> and *p*<sub>*Y*</sub>, the projection of *X* and *Y* onto ℝ<sup>*n*-1</sup> along the *k*<sup>th</sup> direction give the same distribution.
[1]: http://en.wikipedia.org/wiki/Brownian_bridge