Let's think about discrete-time white noise...
For the increments to be independent, you need Z(n) - Z(n-1) to be independent of Z(0), Z(1), ... Z(n-1). The problem is Z(n) - Z(n-1) is not independent of Z(n-1).
You can think about the covariance to help you see this:
cov(Z(n) - Z(n-1) , Z(n-1)) = cov(Z(n), Z(n-1)) - cov(Z(n-1), Z(n-1))
= 0 - sigma^2
Since the cov is not equal to 0, Z(n)-Z(n-1) and Z(n-1) are not independent.
As for the Markov property, it holds in a trivial way. Since the Z's are indep, we have:
P(Z(n) = z(n) | Z(n-1) = z(n-1), Z(n-2) = z(n-2), ...., Z(0) = z(0))
= P(Z(n) = z(n))
and also:
P(Z(n) = z(n) | Z(n-1) = z(n-1)) = P(Z(n) = z(n))
So P(Z(n) = z(n) | Z(n-1) = z(n-1), Z(n-2) = z(n-2), ...., Z(0) = z(0))
= P(Z(n) = z(n) | Z(n-1) = z(n-1))
(they are both just equal to the unconditional probability)
Continuous-time white noise can be thought about in the same sort of way, but you have to use the filtration stuff when you think about all the past values since you can't list them all.