Prove that $\forall d > 0$ there exists $x_1, x_2 \in \mathbb{R}$ such that $x_1 - x_2 = d$ and $f(x_1) = f(x_2)$

Since $f$ is not assumed differentiable you cannot use Rolle's theorem; but the IVT is available, and does the job.

After substracting a suitable constant from $f$ (which is irrelevant for the problem at hand) we may assume that $\lim_{x\to-\infty}f(x)= \lim_{x\to\infty}f(x)=0$, and that there is an $a\in{\mathbb R}$ with $f(a)>0$. Using a standard argument we then can infer that there is a $\xi\in{\mathbb R}$ with $$f(x)\leq f(\xi)\qquad(-\infty<x<\infty)\ .$$ Given a $d>0$ put $g(x):=f(x)-f(x+d)$. Then $$g(\xi-d)=f(\xi-d)-f(\xi)\leq0,\qquad g(\xi)=f(\xi)-f(\xi+d)\geq0\ .$$ As $g$ is continuous it follows that there is a point $x_1\in[\xi-d,\xi]$ with $g(x_1)=0$. Put $x_1+d=:x_2$; then $|x_2-x_1|=d$ and $$f(x_1)-f(x_2)=f(x_1)-f(x_1+d)=g(x_1)=0\ .$$

Fix $d>0$ and put $g(x) = f(x+d)-f(x)$. If for all $x \in \Bbb R$, $g(x) \neq 0$, then $g(\Bbb R) \subset (-\infty,0)\cup(0,\infty)$.

Let $A=g^{-1}((-\infty,0))$ and $B=g^{-1}((0,\infty))$. Assume that neither is empty. Note that $A\cup B = \Bbb R$, and $A$ and $B$ are both open (by continuity of $g$) and that they are disjoint. This contradicts the connectedness of $\Bbb R$. Therefore $A=\emptyset$ or $B = \emptyset$, i.e. $g(x) < 0$ for all $x\in \Bbb R$ or $g(x) > 0$ for all $x\in \Bbb R$.

Now assume $g(x)>0$ for all $x$, then $f(x+d)>f(x)>f(x-d)$ for all $x$. Then, for any fixed $x$,

• $f(x)>f(x-d)>f((x-d)-d) = f(x-2d)>\cdots > f(x-nd)$, for all $n \in \Bbb N$

and

• $f(x)<f(x+d)<f((x+d)+d) = f(x+2d) < \cdots < f(x+nd)$ for all $n \in \Bbb N$

i.e. $f(x+nd)>f(x)>f(x-nd)$ for all $n \in\Bbb N$. Taking $n\to \infty$, we get by continuity that $f(x) = L$. Since $x$ was arbitrary, we get $f(x) = L$ for all $x$. So $g(x) = 0$ for all $x$, contradiction.

Similarly, if $g(x)<0$ for all $x$, we get a contradiction.

We conclude that there exists $x_0$ such that $g(x_0)=0$, i.e. $f(x_0+d) = f(x_0)$, as desired.

Added later on

Looking at it now, there is no need for all of that juggling to prove that $g(x) > 0$ for all $x$ or $g(x) < 0$ for all $x$. Just argue that if at some two points $g(x_1)\le 0$ and $g(x_2)\ge 0$, then by IVT $g(x_0)=0$ somewhere.