I have found several partial answers to that question, all of which used more sophisticated mathematics than what I think is warranted by the question. Now, this may well be because my proofs below are not correct -- and so I thought to write them here and ask for feedback.
First, suppose that we have a sequence $T_n \rightarrow T$, such that for any $n$, we have $f(x + T_n) = f(x)$, where $f$ is a continuous function, and $x$ is any value in its domain. Then we must also have $f(x + T) = f(x)$. To see why this is, let $x$ be fixed; it is immediate that the sequence $x + T_n$ converges to $x + T$. Now, on the one hand, the sequence $f(x + T_n)$ is constant---all its terms are equal to $f(x)$. On the other, because $f$ is continuous, we have $\lim_{n \rightarrow \infty} f (x + T_n) = f (\lim_{n \rightarrow \infty} x + T_n) = f (x + T)$. Thus we conclude that $f(x + T) = f(x)$.
But what happens when $T = 0$? By the above result, we are left with a rather trivial condition, that is true for any function, namely that $f (x + 0) = f(x)$. The next result, however, shows that for a continuous function, if it admits arbitrarily small periods (recall that a period is by definition positive), then it must be a constant function.
Proposition. Let $f$ be a continuous function, for which there exists a sequence $T_n' \rightarrow 0$ of nonzero terms, such that for any $n$, we have $f(x + T_n') = f(x)$. Then $f$ is a constant function.
Proof. It is easy to show that if $f (x + t) = f(x)$ holds, then also $f (x + kt) = f(x)$ holds, for any integer $k$, including negative values (note that the case $k = 0$ is trivial). Thus, we can construct a strictly decreasing sequence $T_n$, composed of only positive terms, such that $f (x + T_n) = f(x)$ for any $n$, and for all $x$ in the domain of $f$, and such that $T_n \rightarrow 0$. (++) Let $a, b$ be two distinct points of the domain of $f$. For any $n$, we can write $b - a = T_n \lfloor (b - a)/T_n \rfloor + r_n$, with $0\le r_n < T_n$ (note this holds regardless of whether $b - a$ is positive or negative). Rearranging, we have $b - r_n = a + T_n \lfloor (b - a)/T_n \rfloor$. As $0\le r_n < T_n$ and $T_n \rightarrow 0$, by the squeeze test we have $r_n \rightarrow 0$. Now, on the one hand, as every $T_n$ is a period of $f$, we have $f(a + T_n \lfloor(b - a)/T_n \rfloor )= f(a)$, which is to say that all the terms of the sequence $f(b - r_n)$ are equal to $f(a)$. But on the other hand, $\lim_{n \rightarrow \infty} b - r_n = b$, and as $f$ is continuous, $\lim_{n \rightarrow \infty} f(b - r_n) = f(\lim_{n \rightarrow \infty} b - r_n) = f(b)$. Hence, we conclude that $f(b) = f(a)$---and as $a$ and $b$ were arbitrary points of the domain of $f$, we conclude that $f$ is a constant function.
Correct, or did I botched up somewhere?
(++): For a brief sketch of how to construct this sequence, let $T_1 = T_1'$, if $T_1'$ is positive, and $-T_1'$ if it is negative. Let $T_i'$ be the next term of sequence $T_n'$ verifying $\lvert T_i' \rvert < \lvert T_1' \rvert$ (one such term can always be found, because $T_n' \rightarrow 0$). Set $T_2 = T_i'$, if $T_i'$ is positive, and $-T_i'$ if it is negative. And so on, and so forth...
Your proof looks perfectly fine to me. Slightly rephrased, you're showing that the set $\{a+kT_n'\mid k\in\mathbb{Z},n\in\mathbb{N}\}$ is dense in $\mathbb{R}$, and since $f$ takes the same value on all points in this set and continuous function are fully determined by their values on a dense set, $f$ must be constant.