This an interesting question.

Think about this way:

W(t)=z1 * sqrt(t),

W(s)=z2 * sqrt(s)

Then

W(t)-W(s)=z1*sqrt(t) – z2*sqrt(s)  is new Brownian and has var = t-s and then  is written as

= z3 * sqrt( t- s)

Where z1, z2 and z3 are different.

Another way: Brownians have been defined for certain properties (including above) and then are proved mathematically that these properties do exists.

Hope this helps.

严格证明是这样的。
为了写得方便，我们假设t>s
var(W(t)-W(s))=E[(W(t)-W(s))^2]-{E[W(t)-W(s)]}^2
=E[(W(t)-W(s))^2]-0
=E[W(t)^2]+E[W(s)^2]-2E[W(t)W(s)]
=t+s-2s
=t-s
其中
E[W(t)W(s)]=E[[W(t)-W(s)+W(s)]W(s)]=E[((W(t)-W(s))W(s)]+E[W(s)^2]
=E[W(t)-W(s)]E[W(s)]+s
=s

LZ's proof basically says that z*sqrt(dt) is NOT Brownian Motion.

hao hao hao hao hao

Normally, using the theory of martingale, it is easy to improve.
And the problem of LZ, I think you had better not take a z normal standard like what you have done. It is not easy to solve those stochastic calculs problem.
Or W(t)=z1*sqrt(t),W(s)=z2*sqrt(s),and z1 z2 are not independent and not the same.
Now E(W(t)|W(s))=W(s) So, E((W(t)-W(s))^2)=E(E((W(t)-W(s))^2|W(s)))=E(W(t)^2-W(s)^2)=t-s
I hope it is clear.
I'm sorry I can not input chinese because I am in the entreprice.