If we construct a square with side length 1, take its diagonal length : $\sqrt{2}$
However I still don't understand HOW it can lie on the number line.
Imagine another irrational number $\pi = 3.1415926535…$
This number is also construct-able, and thus lies successfully on the number line.
However, for such a number to "lie" on the number line, it should be stationary.
If we look at the number $\pi$
$\pi = 3.1…$ We know $3.1 < \pi <3.2$
$\pi = 3.14…$ We know $3.14 < \pi <3.15$
$\pi = 3.141…$ We know $3.141 < \pi <3.142$
$\pi = 3.1415…$ We know $3.1415 < \pi <3.1416$
This list would go on forever, and the number $\pi$ is always "in-between" the two values,
$a < \pi < b$
and thus always in an increment (non-stationary) of a tiny value, compared to $a$.
Best Answer
Your question basically boils down to how do we know that the real numbers are complete, because like you said you can come up with smaller and smaller intervals with rational endpoints that contain your irrational number. If the real numbers are complete, then the intersection of all these intervals will be a single point, your number. There are multiple formal constructions of the real numbers starting with the rationals (Dedekind cuts are one) and you can prove the real numbers are indeed complete based on the construction. So all your limits of convergent sequences exist as a real number.