Confusion on inequality in proof

Question

We are tasked to prove the following: $$\forall x \in\mathbb R \forall\epsilon > 0 \exists \delta > 0 \forall y \in\mathbb R, |x − y| < \delta : |x^2 − y^2| < \epsilon.$$

In the solution, the lecturer has included some rough work:
"We may express $|x^2-y^2|$ as $|x-y| \cdot |x+y|$. We know that $|x-y| < \delta$.
Now, $|x+y| \le |x|+|y| \le |x| + |x| +\delta=2|x|+\delta$"

Why is it that $|x|+|y| \le |x| + |x| +\delta$? I can't see how he deduced this.

Answer 1

We have $|x-y|<\delta$.

$$|y|=|y-x+x|\leq |x-y|+|x| <\delta +|x|$$

Answer 2

That is because the possible value of $y$ will be in $(x-\delta,x+\delta)$ since $|x-y|<\delta$.

Answer 3

Let $|x - y| < \delta$. Then, $$ x - \delta <y<x + \delta\\ |y| < \max\{| x - \delta|,|x+\delta|\} \le |x| + \delta $$ Therefore, $$ |y| - |x| < \delta $$