If $A\subseteq\mathbb{R}$, $A$ dense in $\mathbb{R}$ and $f$ and $g$ are two continuous functions on $\mathbb{R}$ such that $f(a)=g(a)\space\forall\space a\in A$, then $f(x)=g(x)\space\forall\space x\in\mathbb{R}$
How can I use the previous statement to determine all functions that satisfy Cauchy's functional equation in $\mathbb{R}$? That is, all functions $f: \mathbb{R}\rightarrow\mathbb{R}$ such that $\forall\space x,y\in\mathbb{R}$, $f(x+y)=f(x)+f(y)$
It seems to me you think about all continuous functions. If $f$ is additive (i.e. $f$ satisfies Cauchy functional equation), then $f(x)=xf(1)$ on rationals. This holds for any sdditive function, continuity is not necessary. Now let $f$ be continuous. Take $g(x)=xf(1)$ for all $x\in\Bbb R.$ Then $g=f$ on rationals, both of them are continuous, so $f=g$ on all reals and $f(x)=xf(1)$ for any $x\in\Bbb R$.
If we are looking for all (not necessarily cotinuous) additive funtions, we need to observe that any such function is a linear map of a linear space $\Bbb R$ over a field $\Bbb Q$ into itself. So, it is uniquely determined by its values on the (Hamel) basis, which does exist if we assume the Axiom of Choice. It is easy to check that an additive function is contnuous (i.e. of the form $f(x)=xf(1)$) iff $f$ restricted to a Hamel basis is constant. So, there are discontinuous additive functions. It could be (not so easily) proved that a graph of a discontinuous additive function is a dense subset of a plane. Then such functions are very strange. Nevertheless, if we assume slight regularity conditions (s.t. measurability, lower or upper semicontinuity at one point, local lower or upper boundedness, monotonicity and so on), then $f$ is necessarily continuous. This implies that a discontinuous additive function is, for instance, non-measurable, non-integrable and so on. For such topics one should read on Bernstein-Doetsch Theorem and its counterparts. A great book on additive functions is this Kuczma's monograph: http://www.springer.com/gb/book/9783764387488.