Others already told you that a proper proof is more involved. However I want to point out, that there is a bit of intuition behind those rules. For me, having seen the way ordinary calculus is treated by physicists, the following is rather insightful:

Think of $dt$ as a really small increase in time $dB_t$ as the change the Brownian motion does in this small time increase and so on. Then the change of a quantity $X_t$ in some time interval is just the sum of all such small changes $dX_t$, or bending notation, the integral. In a way this is also what you do when defining the stochastic integral.

Using this, (dt)^2 is simple. If you separate your interval in about $\sim N$ steps, then $dt \sim \frac{1}{N}$. So $(dt)^2 \sim \frac{1}{N^2}$. Yet if you sum $N$ steps of size $\frac{1}{N^2}$, in the limit $N\to \infty$, you will end up with nothing. So $(dt)^2$ has no effect.

Now $dB_t$ is a strange beast. Comparatively it is of size $\sqrt{dt}$ (have a look at the scaling of the probability density in $x$ and $t$), so by the reasoning above, the integral should explode. Yet it changes sign all the time and those changes cancel each other quite nicely in the end.

On the other hand $(dB_t)^2 \sim \sqrt{dt}^2 = dt$ and has a fixed sign, so the calculation rule at least is not surprising. (That it is actually equal, however, is a not so trivial matter.)

The last one, $dB_t dt$ again is simple, as this is of order $\sim (dt)^{3/2}$, which allows us to use a similar argument to $(dt)^2$.

I should again stress, that this is only intuition, which cannot prove anything, only help in finding what to prove. For example using this, Ito's formula looks just like a Taylor expansion in $dB_t$ and $dt$, where you throw away the terms of order larger than $dt$.