Prove that $|x − y| < \epsilon$ , $|y − z| < \epsilon$ implies $|x-z|<2\epsilon$

Question

I need the math.triangle inequality formula, but I still didn't get it fully. be die|x|+|y| <= |x+y|

|x|+|y| <= |x+y| I put in the values

|x-y| < ε <= |x|-|y| < ε |y-z| < ε <= |y|-z| < ε

|x-z| <2ε

|x-z| <2ε <= |x| -|z| <2ε
But that still doesn't prove much I guess. I really don't know and spent quite some time on this sry my english isn't so good!

Answer 1

$$|x - z| = |(x-y) + (y-z)| \leq |x - y| + |y - z| < \epsilon + \epsilon$$

Answer 2

$|x-y|< \epsilon \rightarrow(1)$

$|y-z|< \epsilon \rightarrow(2)$

By $(1)+(2)$ we get that $|x-y|+|y-z|<2\epsilon$

By triangle inequality we have $|a+b|\leq|a|+|b|$ . So put $a=x-y $ and $b=y-z$

Therefore $|(x-y)+(y-z)|\leq |x-y|+|y-z|<2\epsilon$.

Thus $|x-z|<2\epsilon$.