This is the proof, which I mostly understand except for one bit:
You have $h_1 \in H_1$ and $h_2 \in H_2$.
We also have $h_1^{-1}(h_2^{-1}h_1h_2) \in H_1$, because $h_2^{-1}h_1h_2 \in h_2^{-1}H_1h_2 = H_1$. Similarly, we have $(h_1^{-1}h_2^{-1}h_1)h_2 \in H_2$. Therefore
$ h_1^{-1}h_2^{-1}h_1h_2 \in H_1 \cap H_2 = \{1_G\} $
and so $h_1^{-1}h_2^{-1}h_1h_2 = \{1_G\}$. Let's first multiply everything on the left by $h_1$
$h_1^{-1}h_2^{-1}h_1h_2 = \{1_G\}$ $ h_1 h_1^{-1}h_2^{-1}h_1h_2 = h_1 \{1_G\}$ $ h_2^{-1}h_1h_2 = h_1 \{1_G\}$
Multiply both sides on the left by $h_2$ giving us
$ h_1h_2 = h_2 h_1 $
The bit I don't get is right at the beginning. Why is this correct: "We also have $h_1^{-1}(h_2^{-1}h_1h_2) \in H_1$, because $h_2^{-1}h_1h_2 \in h_2^{-1}H_1h_2 = H_1$."