Variance of the Wiener Increment

I have a Wiener process $$W(t)$$, which is a normally distributed random variable with mean $$\langle W(t)\rangle = \mu = 0$$ and variance $$\langle W(t)^2\rangle = \sigma^2 = t$$. The angled brackets $$\langle\ \rangle$$ indicate an average over all realisations of the Wiener process.

The Wiener increment $$\Delta W$$ is defined as:

$$\Delta W(t) = W(t + \Delta t) - W(t)$$

which corresponds to the time increment $$\Delta t$$. The mean of $$\Delta W$$ is zero, since the means of both $$W(t + \Delta t)$$ and $$W(t)$$ are zero.

I am trying to derive the variance of $$\Delta W$$, which is $$\langle (\Delta W)^2\rangle$$. So far I have:

$$(\Delta W)^2 = W(t + \Delta t)^2 + W(t)^2 -2W(t + \Delta t)W(t)$$

$$\langle (\Delta W)^2\rangle = \langle W(t + \Delta t)^2\rangle + \langle W(t)^2\rangle -2\langle W(t + \Delta t)W(t)\rangle$$

$$\langle (\Delta W)^2\rangle = t + \Delta t + t -2\langle W(t + \Delta t)W(t)\rangle$$

I am not sure how to evaluate the last term (or if what I have done so far is correct) and would appreciate some help. I have been told that the correct answer is $$\Delta t$$ and am trying to verify this. Many thanks!

So you can take a shortcut if you know that the Wiener process has independent increments. This is stronger than merely being Markov, because it follows that the distribution of the increments starting from a time $$t$$ do not depend even on $$W_t$$. Once you know that (which in some definitions of the Wiener process is built into the definition), it follows that the distribution and thus moments of $$\Delta W$$ do not depend on $$W$$ at the initial time. Thus in particular moments of $$W_t - W_s$$ for $$t \geq s$$ are just the same moments of $$W_{t-s}$$.
But the alternative is to show that the covariance function of the Wiener process is $$E[W_t W_s] = \min \{ t,s \}$$. Exactly how to do this will depend on what you already have available to you to use.