How does this proof of Theorem 1 in Spivak's Calculus work?

Hello I just started Spivak's Calculus and I've come across something I don't really understand. What is trying to be proved is $$ |a+b|\leq|a|+|b|\,. $$ The proof is based on the observation that $|a|=\sqrt{a^2}$. So: $$ \begin{align*} (|a+b|)^2 = (a+b)^2 &= a^2+2ab+b^2 \\ &\leq a^2+2|a|\times|b|+b^2 \\ &= |a^2|+2|a|\times|b|+|b^2| \\ &= (|a|+|b|)^2 \\ \end{align*} $$ Now what I don't understand is the change in relationship from line 1 to 2 ($= to \leq$) and from line 2 to 3 ($\leq to =$).

I attach the page because it's clearer there (I highlighted the proof in yellow). Actual proof from the book

The lines are read left to right and top to bottom. So the symbol ≤ on line 2 relates the last expression of line 1 to the next expression on line 2. Similarly, the symbol = on line 3 relates the last expression of line 2 to the next expression on line 3. The same thing could be written in one line as follows: \begin{equation} \begin{split} (|a+b|^2) = (a+b)^2 &= a^2+2ab+b^2 \leq a^2+2|a|\times|b|+b^2 = |a^2|+2|a|\times|b|+|b^2| = (|a|+|b|)^2, \end{split} \end{equation} but it is easier on the eyes to write each new expression on a new line.

For any number $x \in \mathbb{R}: x \leq |x|$

Hence, $ab \leq |ab| = |a||b| \implies 2ab \leq 2|a||b| \implies a^2 + 2ab + b^2 \leq a^2 + 2 |a||b| + b ^2$

$$a^2+2ab+b^2 \leq a^2+2|a|\times|b|+b^2 $$ because $$ab \le |a||b|$$

Also $$a^2+2|a|\times|b|+b^2 = |a^2|+2|a|\times|b|+|b^2| $$

because $a^2 = |a|^2 $ and $b^2 = |b|^2 $