Here is a sketch of a proof (it uses axiom of choice).
Let's introduce a equivalence relation on real numbers.
We say that two numbers $x$ and $y$ are "equal" of there exists a finite
set of reals $a_0 = x$, $a_1$, $a_2$, $\cdots$, $a_{n-1}$, $a_n = y$ such
that for any $i$ we have $a_i \in \Phi(a_{i+1})$ or $a_{i+1} \in \Phi(a_i)$.
Note that $n$ could be zero, so $x$ and $x$ are also considered equal.
It's clear that this equivalence relation is symmetric, reflexive and transitive.
Hence we can consider equivalence classes under this relation.
Now, every equivalence classes is countable (because $\Phi(x)$ is finite for all $x$).
Hence we have splitted reals into a set of disjoint countable sets.
This implies that number of such sets is a continuum.
Now we can just take any element from each set to form $S$.