Let $\{x\}=x-\lfloor x\rfloor$, and let $1-\{-x\}=f(x)$. Then let $J=[x, y]$. (If there is an open interval, then change the necessary brackets to parentheses.) We have that $T(J)=[f(x-a), f(y-a)]$, "wrapping around" from 1 to 0 if necessary. Proof: $T(J)$ takes a point on $J$ and subtracts $a$ from it, and if it is less than or equal to 0 adds 1 to it. So it is almost the function $T(x)=\{x-a\}$, but instead of $T(a)=0$, $T(a)=1$. The rewrite of $f(x)$ takes care of that. Corollary: $T^n(J)=[f(x-an), f(y-an)]$. Because $T(J)$ admits an inverse, we can say: Corollary: $T^{-1}(J)=[f(x+a), f(y+a)]$, and thus $T^{-n}(J)=[f(x+an), f(y+an)]$. WLOG, we need to only show the conclusion for a continuous interval with no breaks. If $J$ is not, we can take any continuous interval and use that. Either $a$ is rational or irrational. If $a$ is rational, say $\frac{b}{c}$, then choosing $n=c$ immediately gives $T^n(J)=J$. If, on the other hand, $a$ is irrational, then let $\left\lfloor\frac{1}{y-x}\right\rfloor+1=m$. The intervals $T^k(J)$, for $k=1, 2, 3\ldots m$ each cover a length of $y-x$, again remembering to wrap around from 1 to 0 if necessary. The total interval is then $m(y-x)>1$. Hence there is some two intervals $T^o(J)$ and $T^p(J)$ that intersect. This means that there is some point $T^p(S)$ that is in both $T^o(J)$ and $T^p(J)$, say $T^p(S)=T^o(U)$. Suppose $o>p$. We can take the inverse of $T$ $p$ times, to give that $T^{o-p}(U)=S$ and thus if $n=o-p$, then $S\in T^n(J)\cap J$ so $T^n(J)\cap J\neq\emptyset$.