Can someone explain this proof of the product property of square roots?

Let $x=\sqrt{a}\cdot\sqrt{b}$. Step $5$ shows that $x^2=ab$. Step $6$ shows that $x\ge 0$. Step $8$ says that there is exactly one number with these properties, and it’s denoted by $\sqrt{ab}$. Since $x$ has these properties, and $\sqrt{ab}$ is a name of the only number with these properties, it must be the case that $x=\sqrt{ab}$, which is what Step $9$ says.

Let's just prove the last statement in your question. Let $a,b \ge 0$ and assume $a^2 = b^2$.

Assume for the sake of contradiction $a \ne b$. WLOG, $a > b$, so $a = b+x$ for some real $x > 0$.

Now $$ a^2 = (b+x)^2 = b^2 + 2bx + x^2 $$ and since $a^2 = b^2$, subtracting it from both sides yields $$0 = 2bx + x^2 = x(2b+x).$$ Hence, either $x=0$ or $x = -2b$. But $b \ge 0$, so if $x = -2b$ then $x \le 0$, which contradicts $x > 0$. Then, $x=0$, which also contradicts $x > 0$.

Hence $a=b$.