I'm not sure whether the domain $\mathbb{R}$ is supposed to be part of the statement to be proved or disproved (usually $\mathbb{R}$ is reserved for denoting the real numbers). That is, whether we are supposed to prove the statement if (1) then (2) for all domains $\mathbb{R}$ or if the domain is given to us.
To make things clear, let's call the statement you are trying to prove or disprove, that is "if (1) then (2)", for $S$.
If the statement $S$ is supposed to hold for arbitrary domains $\mathbb{R}$, this is a very general condition that needs to be satisfied and thus one can imagine that the statement $S$ might be false. So one can first try to come up with a counterexample to the statement $S$ by finding a domain $\mathbb{R}$ that makes the statement false. Indeed, any domain $\mathbb{R}$ with a single element is a counter example to the statement $S$.
Now, I suppose it is more likely that the domain $\mathbb{R}$ is given to us. Let us for the sake of discussion take it to denote some subset of the real numbers. You write:
I'm not looking for an explicit answer, just some tips or ways to approach this type of problem where there are so many if-then statements and things to consider. I just want to know how to interpret these types of multi-layered problems.
If there seems to be too much at once to consider, try to break the statement $S$ down and consider only some parts of it at once. In this example, $S$ is of the form
$$\exists a \in \mathbb{R} : P_1(a) \quad (1.) \implies
\forall a' \in \mathbb{R} : P_2(a') \quad (2.)$$
Where $P_1(a)$ and $P_2(a')$ simply denotes the rest of (1.) and (2.) which we don't care about at the moment. I use the variables $a'$ and $b'$ in (2.) rather than $a$ and $b$, in order to not mix up the variables in (1.) and (2.).
The statement $S$ says that if there is some element in $\mathbb{R}$ such that if the proposition $P_1$ is true then the proposition $P_2$ is true for every single element in $\mathbb{R}$. This seems like a rather strong condition to fullfill, but not impossible. In order to reduce the amount of logical symbols we can simply decide to call this $a$ in (1.), that is assumed to exist, $a_0$.
Now lets consider a larger chunk of (1) and (2):
\begin{align}
&\forall b \in \mathbb{R} : a_0 < b \implies P(a_0, b)\\
& \implies \\
&\forall a' \in \mathbb{R}, \exists b' \in \mathbb{R} \text{ such that } a' < b' \text{ and } P(a', b')
\end{align}
Here we see that the statement $S$ makes a rather strong assertion, because $(1.)$ only says something about $a_0$ and all elements larger than $a_0$; it does not say anything explicitly about elements smaller than $a_0$.
The analysis so far would at least make me inclined to try to come up with a counter example to the statement.
I don't know how to define the functions such that any two functions $f$, $g$ obey $∃a∈ℝ$ s.t. $∀b∈ℝ \,\,a<b \implies f(a)-g(a)<f(b)-g(b).$ Any ideas? I could provide what I have tried but it does not amount to much.
You can start by just looking at the part that says
$$ a < b \implies f(a) - g(a) < f(b) - g(b) \quad (*)$$
If it feels like to much to come up with two functions at once, one strategy you can try is to just pick one function for, say, $g(x)$ kind of arbitrarily. Say you come up with $g(x) = x$. Now, it might help to draw $g(x)$ on paper. See if you can now find/draw $f(x)$, such that as $x$ becomes larger, $f(x) - g(x)$ also becomes larger (that is $f(x)$ increases faster than $g(x)$). It might help to assume $f(x)$ is never less than $g(x)$. If you can do that, then you have found one pair of functions that satisfy $(*)$ and thus you can pick any $x$ as your $a$ in $(1.)$.
I guess it's worth pointing out that in order to find a counter example to the statement $S$, however, more care is needed. We would like to make $(2.)$ false while at the same time $(1.)$ being true. One way to make $(2.)$ false is if there is some $a'_1$ such that $f(a'_1) - g(a'_1) \geq f(x) - g(x)$ for all $x > a'_1$. This means we can not let the difference $f(x) - g(x)$ grow without bound as we can in order to just satisfy $(1.)$. But $(1.)$ does not require that the difference $f(x) - g(x)$ grows without bound, so it should be possible to define $f$ and $g$ in ways that make $(2.)$ false while $(1.)$ still is true, which would provide a counter example to the statement $S$. But I guess these considerations might be to explicitly related to the statement at hand, I just thought it was worth pointing out.
