For two geometric random variables $X_t$ and $X_{t+1}$. Consider a coupling of $(X_t, X_{t+1})$. Let $\{U_i\}_{i\geq 1}$ be a sequence of iid random uniform on $[0,1]$.
How to show that $X_{t+1}-X_t$ is independent of $X_t$?
I guess $$P(X_{t+1}=X_t|X_t)=\frac{t}{t+1}.$$
Just show that $P(X_{t+1}-X_t=d\mid X_t=i)$ is free of $i$ for every $d$. Since you've verified this for $d=0$, and since $X_{t+1}\ge X_t$, it remains to check the case $d\ge1$. For this case you can write out $$ P(X_{t+1}-X_t=d\mid X_t=i)=\\ \textstyle P\left(U_i>\frac1{t+1}, U_{i+1}>\frac1{t+1},\ldots,U_{i+d-1}>\frac1{t+1},U_{i+d}\le\frac1{t+1}\ \big|\ U_1>\frac1t,\ldots,U_{i-1}>\frac1t,U_i\le\frac1t\right) $$ By independence of the $U$'s this reduces to $$\textstyle P(\underbrace{U_{i+1}>\frac1{t+1},\ldots,U_{i+d-1}>\frac1{t+1}}_{\text{$d-1$ conditions}},U_{i+d}\le\frac1{t+1})P\left( U_i>\frac1{t+1}\ \big | \ U_i\le\frac1t\right) $$ which is clearly the same value for any $i$.
EDIT: Why is it enough to show that $P(X_{t+1}-X_t=d\mid X_t=i)$ is free of $i$? If $P(X_{t+1}-X_t=d\mid X_t=i)$ is a constant $c_d$ not depending on $i$ then $$ P(X_{t+1}-X_t=d, X_t=i)=P(X_{t+1}-X_t=d\mid X_t=i)P(X_t=i)=c_d P(X_t=i).$$ Summing over $i$ we obtain $$P(X_{t+1}-X_t=d)=\sum_i P(X_{t+1}-X_t=d, X_t=i)=c_d\sum_iP(X_t=i)=c_d $$ and we conclude $$P(X_{t+1}-X_t=d, X_t=i)=P(X_{t+1}-X_t=d)P(X_t=i) $$ which implies independence. Note that it is not necessary to find the value of $c_d$, although in this case it is equal to $$ \left(1-\frac1{t+1}\right)^{d-1}\frac1{t+1}P\left( U_i>\frac1{t+1}\ \big | \ U_i\le\frac1t\right) $$ where $$P\left( U_i>\frac1{t+1}\ \big | \ U_i\le\frac1t\right)=\frac{P(\frac1{t+1}<U\le\frac1t)}{P(U\le \frac1t)}=\frac{\frac1t -\frac1{t+1}}{\frac 1t}=\frac 1{t+1}. $$