Suppose f is continuous at c. Prove abs(f) is continuous at c.

Notice that for any real numbers a and b, the triangle inequality gives us:
|a| <= |a - b| + |b|, and
|b| <= |b - a| + |a|

Those can be rewritten as:
|a| - |b| <= |a - b|
|b| - |a| <= |a - b|

Combining the two gives the following FACT:
| |a| - |b| | <= |a - b|

Since f is continuous at c, then for any ε>0, there exists a δ>0 such that:
|f(x) - f(c)| < ε whenever |x - c| < δ.

Hence, if we substitute a = f(x) and b = f(c) in the above FACT, we can conclude that:
| |f(x)| - |f(c)| | < ε whenever |x - c| < δ.

And that shows |f| is continuous at c.