How to show that such a sequence exist? Suppose that a non-constant function $\gamma : \mathbb{R} \to \mathbb{R}$ is $T$ -periodic for some $T \neq  0$. This exercise shows that there is a smallest positive number $T_0$ such that $\gamma$ is $T_0$-periodic. The proof uses a little real analysis.
Suppose for a contradiction that there is no such $T_0$.


*

*Show that there is a sequence $T_1, T_2, T_3, \dots$ such that $T_1 > T_2 >
T_3 > \dots > 0$ and that $\gamma$ is $T_r$-periodic for all $r \geq 1$.

*Show that the sequence $\{T_r\}$ of $(1.)$ can be chosen so that $T_r \to 0$
as $r \to \infty$.

*Show that the existence of a sequence $\{T_r\}$ as of $(1.)$ such that
$T_r \to 0$ as $r \to \infty$ implies that $\gamma$ is constant. 



At the First step I am facing difficulties to show that such a sequence exist. 
Any hints how to do that? 
 A: If there is no such $T_0$, by definition that means that for every $T_0 > 0$ there exists $0 <T^\prime <T_0$ such that $\gamma$ is $T^\prime$-periodic.
Use this to define the sequence iteratively: to get $T_1$, take $T_0=1$ in the above, and let $T_1$ be the corresponding $T^\prime$. Then, "take $T_0=T_1$" in the above, and let $T_2$ be the new "corresponding $T^\prime$. Etc.: given $T_n$, define $T_{n+1}$ to be the "$T^\prime$" corresponding to "$T_0=T_n$."

 For the second question, one "trick" is to define the sequence above by ensuring it decreases at least by a factor $2$ every time. I.e., take $T_{n+1}$ to be the $T^\prime$ corresponding to "$T_0=\frac{T_n}{2}$", not to "$T_0=T_n$".

Edit: additionally, as André Nicolas points out, you will not be able to prove 3. without some additional assumption on $\gamma$. Indeed, as stated it is false -- there are non-constant functions without a smallest positive period.
A: We concentrate on 2). Suppose that $\gamma$ is periodic but has no smallest positive period. We show that there is a sequence of periods of $\gamma$ with limit $0$. 
Let $W$ be the infimum of all the periods. Suppose that $W$ is positive, and $W$ is not a period. Then there exists a strictly decreasing sequence of $W_1,W_2,W_3,\dots$ of periods, with limit $W$. In particular, for any integer $m$ there exist periods $W_i\lt W_j$ such that $W_j-W_i\lt \frac{1}{m}$. But $\gamma(x+W_j-W_i)=\gamma(x-W_i)=\gamma(x)$, We conclude that for any positive integer $m$, $\gamma$ has a positive period $\lt \frac{1}{m}$.
So we have shown that if there is no smallest positive period, there is sequence of periods with limit $0$.
For 3), as pointed out in a comment, we will need to make a further assumption on $\gamma$. Continuity will do it.
Added: We show that if there is a point $a$ at which the function $\gamma(x)$ is continuous, then $\gamma$ is a constant function. We include more or less full detail.
Suppose that $\gamma(a)=c$, and that there exists a $b$, and a $d\ne c$ such that $\gamma(b)=d$. We will derive a contradiction. It then will follow that $\gamma(x)=c$ for all $x$.
By the result of 2), there is a strictly decreasing sequence $(W_i)$ of periods, with limit $0$. In particular, we may choose the $W_i$ so that $W_i\lt \frac{1}{i}$ for any positive integer $i$.
Since $\gamma(b)=d$, we have $\gamma(b+kW_i)=\gamma(b)=d$ for every integer $k$, positive, negative, or $0$. Since $W_i\lt \frac{1}{i}$, it follows that for any positive integer $i$ there is a $b_i$ such that $|b_i-a|\lt \frac{1}{i}$ and $\gamma(b_i)=d$.
So we have found a sequence $(b_i)$ such that $\lim_{i\to\infty} b_i=a$ and $\gamma(b_i)=d$ for all $i$. Since $f(a)=c\ne d$, this contradicts the continuity of $\gamma(x)$ at $x=a$. 
