How to write a proof that holds for any two functions that meet some condition which implies another condition? I have no idea how to start a proof when given something along the lines of:

If $f$ and $g$ are any two functions that have some domain
$\mathbb{R}$ AND obey the property
$$\exists a \in \mathbb{R} \text{ such that } \forall b \in
 \mathbb{R}, \;\;\;a<b \implies f(a)-g(a)<f(b)-g(b),\quad\quad\;\;\;
(1)$$
THEN that implies it also obeys the following:
$$\forall a \in \mathbb{R},  \exists b \in \mathbb{R} \text{ such that
 }  \;\;a<b \,\text{ AND }\, f(a)-g(a)<f(b)-g(b). \quad\quad\quad (2)$$
In other words, IF $f$ and $g$ are any two functions where (1)
holds true for any $f$ and $g$, then the same $f$ and $g$ also true for (2), but the a and b values can change.

So, both $f$ and $g$ must be in the given domain and have the property (1). Given this, how do you prove or disprove that $f$ and $g$ must also have the property given by (2)?
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. I have asked at my school but those explanations just confuse me more. 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.
 A: $$\exists a \in \mathbb{R}  \;\;\forall b \in \mathbb{R} \;\;\big(a<b \implies f(a)-g(a)<f(b)-g(b)\big)\tag1$$
$$\forall a \in \mathbb{R}  \;\;\exists b \in \mathbb{R} \;\;\big(a<b \,\text{ AND }\, f(a)-g(a)\geq f(b)-g(b)\big)\tag{$\lnot$1}$$
$$\forall a \in \mathbb{R} \;\;\exists b \in \mathbb{R}  \;\;\big(a<b \,\text{ AND }\, f(a)-g(a)<f(b)-g(b)\big) \tag2$$
$$\exists a \in \mathbb{R} \;\;\forall b \in \mathbb{R}  \;\;\big(a<b \implies f(a)-g(a)\geq f(b)-g(b)\big). \tag{$\lnot$2}$$
$h(x):=f(x)-g(x).$


*

*We want to prove that $$\exists f,g\text{ on }\mathbb R \,\big((1)\implies(2)\big),$$ which is equivalent to $$\exists f,g\text{ on }\mathbb R \,\big(\lnot1) \,\text{ OR }\, (2)\big),$$ which is equivalent to $\quad\exists h\text{ on }\mathbb R\;\;\forall x,y \in \mathbb{R}  \;\;\exists b \in \mathbb{R},$ $$\big(x<b \,\text{ AND }\, h(x)\geq h(b)\big) \,\text{ OR }\, \big(y<b \,\text{ AND }\, h(y)<h(b)\big),$$ which is a consequence of $\quad\exists h\text{ on }\mathbb R\;\;\forall x,y \in \mathbb{R}  \;\;\exists b \in \mathbb{R},$ $$\big(x<b \,\text{ AND }\, h(x)\geq h(b)\big).$$
Put $h(x)\equiv0.$



*We want to prove that $$\exists f,g\text{ on }\mathbb R \,\big((1)\kern.6em\not\kern -.6em\implies(2)\big),$$ which is equivalent to $$\exists f,g\text{ on }\mathbb R \,\big(1) \,\text{ AND }\, (\lnot2)\big),$$ which is equivalent to $\quad\exists h\text{ on }\mathbb R\;\;\exists x,y \in \mathbb{R}  \;\;\forall b \in \mathbb{R},$ $$\big(b>x \implies h(b)>h(x) \big) \,\text{ AND }\, \big(b>y \implies h(b)\leq h(y)\big).$$
Put \begin{align}h(x) &= \begin{cases}|x| &\text{ if }x<1; \\1 &\text{ if }x\geq1,\end{cases}, \,x=0, \,y=-1.\end{align}

P.S. The above choices of $h(x)$ were provided by Paradox. This Answer has pretty much been written collaboratively with them.
A: 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.
