Let $S =\left\{\ f : \mathbb R \to \mathbb R | \exists \in > 0 : \forall δ > 0, |x-y| < δ \implies |f(x)-f(y)| < \in \right\}$ then
$S = \left\{f : \mathbb R \to \mathbb R | f\text{ is continuous}\right\}$
$S = \left\{f : \mathbb R \to \mathbb R | f\text{ is uniformly continuous}\right\}$
$S = \left\{f : \mathbb R \to \mathbb R | f\text{ is bounded}\right\}$
$S = \left\{f : \mathbb R \to \mathbb R | f\text{ is constant}\right\}$
My Attempt : Let $y = a$ (constant) and $x \in \Bbb R$ then $|f(x)-f(a)| < \in$ $\implies |f(x)|-|f(a)| < \in$ $\implies |f(x)| < \in + |f(a)|$ $\implies$ f is bounded. Am I right ? How to discard other options. I'm confusing with the fact that if $y = a$ (constant) then $f(a)$ finitely exist ( why ? ).
Your attempt is right but I would have worded it differently.
We are going to prove that every function in $S$ is bounded and that every bounded function is in $S$. Thus the answer is 3.
Let $f\in S$, then There exist an $\epsilon>0$ such that for every $\delta>0$, $$\lvert x-y\rvert<\delta\implies\lvert f(x)-f(y)\rvert<\epsilon$$ Since it is true for every $\delta$, we could take it as large as we want. If we fix the value of $y=a\in\Bbb R$, then we must have that for every $x\in\Bbb R$ $$\lvert f(x)-f(a)\rvert<\epsilon$$ Since $y=a$ is part of the domain of the function, then $f(a)$ exist. If we remove absolute value, we have $$-\epsilon <f(x)-f(a)<\epsilon$$ Adding $f(a)$ to every term, we gave $$f(a)-\epsilon<f(x)<f(a)+\epsilon$$ Thus $f$ is bounded.
Conversly, if $f$ is bounded, then there exist $M, N\in\Bbb R$ such that $$N\leq f(x)\leq M\qquad\forall x\in\Bbb R$$ Let take $$\epsilon=2\times \max(|M|, |N|)+1$$ Then for every $x, y\in\Bbb R$, we have $$\lvert f(x)-f(y)\rvert\leq \lvert f(x)\rvert+\lvert f(y)\rvert\leq \max(|M|, |N|)+\max(|M|, |N|)=2\times\max(|M|, |N|)<\epsilon$$ Then $f\in S$.
Conclusion, your definition of $S$ is equivalent to 3.
For the other options, if $f$ is constant, it is bounded, so it is include in case 3.
For continuous or uniformly continuous function, we only need an example of function that doesn't respect the condition.
Let $f(x)=x$. Obviously, $f$ is uniformly continuous, but does it satisfy the condition to be in $S$?
Let $\epsilon>0$, take $y=x+\epsilon+1$. There exist a $\delta>0$ such that $\lvert x-y\rvert<\delta$, so $\lvert f(x)-f(y)\rvert$ must be smaller than $\epsilon$. But $$\lvert f(x)-f(y)\rvert=\lvert x-(x+\epsilon+1)\rvert =\lvert-\epsilon-1\rvert=\epsilon+1>\epsilon$$
Conclusion, for $f(x)=x$, it is impossible to find the $\epsilon$ required by the definition of $S$. We found an uniformly continuous function not in $S$, Thus $S$ is not equivalent to 1 or 2.