If you want to show that is continuous at , you should split according to which is smaller at , first. But this minimizer may change when we pick in a neighbourhood of .

Let . There is such that implies both and . (By the way, the numbering of the functions is a bit strange)

Assume , so that . Then, if , we have .

And we have both and , and one of these (depending if or the contrary) gives the inequality .

Another quick proof relies on the identities : and , once you know that the absolute value of a continuous function is continuous (this results easily from the triangular inequality).