By choosing $a = 2$, $c = 2$, the author demonstrates that the equation can be satisfied when $a \neq 0$ since
$$(a + c)(a - c) = (2 + 2)(2 - 2) = 4 \cdot 0 = 0$$
Hence, $a = 2$, $c = 2$, is a counterexample to the claim that $a = 0$ must be true. There are others. For instance, $a = 1$, $c = 1$ or $a = 1$, $c= -1$. However, only one counterexample is required to demonstrate that a statement is false.
What the author is stating is that $a = 2$, $c = 2$ is also a counterexample to the claim that $b = 0$ must be true or that $a = -c$ must be true for the equation $(a + c)(a - c) = 0$ to be satisfied.
Similarly, if $a = 2$ and $c = -2$, then
$$(a + c)(a - c) = [2 + (-2)][2 - (-2)] = 0 \cdot 4 = 0$$
so $a = 2$, $c = -2$ is a counterexample to the claim that $a = c$ must be true for the equation $(a + c)(a - c) = 0$ to be satisfied. Again, there are other counterexamples such as $a = 1$, $c = -1$.
As for the remaining choice, since $(a + c)(a - c) = a^2 - c^2$, $$(a + c)(a - c) = 0 \Rightarrow a^2 - c^2 = 0 \Rightarrow a^2 = c^2$$