Real Analysis - continuous extension function

Let f : Q → R be a uniformly continuous function. Show that there exists a
uniformly continuous function f : R → R such that f (x) = f (x) for all x ∈ Q. (Notation: bold f is the function f extended to)

Using the theorem (a function with open interval (a,b) can be uniformly continuous iff the limits La=lim f (x) as x->a, Lb=lim f (x) as x->b, exist and the extended function f, with closed interval [a,b], is defined by f(x), La, and Lb ), I know how to prove uniformly continuity of functions with concrete intervals but I don't really know how to approach this question. Should I also demonstrate the limits exist and find limits using Cauchy sequence? I don't know what exactly I show do. Please help me here. Thank you!