No, there exist such functions that don't have a minimal period.
Set $p_0=1$ and choose $p_n \in \mathbb Q, n=1,2,\ldots$ with
$$ p_1 > p_2 > \ldots > p_n > p_{n+1} > \ldots,\quad\lim_{n \to \infty}p_n=1=p_0.$$
Those $p_n, n=0,1,\ldots$ will be the periods of a function that is very close to the final function $f$ that will serve as example.
For each $s \in \mathbb R$, define
$$D_s=\left\{x \in \mathbb R: x=s+\sum_{n=0}^{\infty}k_np_n, k_n \in \mathbb Z, k_n \ge 0, n=0,1,\ldots\right\} \tag{1}, \label{DefDs}$$
Since $\forall n \ge 0: p_n \ge 1, k_n \ge 0$, the infinite sum in \eqref{DefDs} is in fact always finite, only a finite number of $k_n$ can be non-zero.
From the definition it follows easily that
$$ \forall n \ge 0: x \in D_s \Rightarrow x+p_n \in D_s \tag{2} \label{RightPeriod}.$$
The main toplogical property of $D_s$ that we need to construct the example $f$ is
Lemma 1: For each $s \in \mathbb R,\, D_s$ is a closed subset of $\mathbb R$.
The proof of Lemma 1 will come at the end, as to not interrupt the construction of the function $f$.
Choose an irrational number $\alpha$ with
$$ 0 < \alpha < \frac12. \tag{3} \label{DefAlpha}$$
We have
$$ D_0 \cap D_\alpha = \emptyset, \tag{4} \label {Disjoint}$$
as otherwise $x \in D_0 \cap D_\alpha$ would imply
$$ 0+\sum_{n=0}^{\infty}k_np_n = \alpha + \sum_{n=0}^{\infty}l_np_n$$
for some $k_n,l_n \in \mathbb Z, k_n,l_n \ge 0$. Since only finitely many $k_n,l_n$ can be non-zero, this would imply that $\alpha$ is a linear combination with integer coefficients of finitely many rational $p_n$, which is contradiction to $\alpha$ being irrational.
Define
$$D^* = D_0 \cup D_\alpha \tag{5} \label{DefDStar},$$
and
$$f^*: D^* \rightarrow \mathbb R, f^*(x)=
\begin{cases}
0, & \text { if } x \in D_0 \\
1, & \text { if } x \in D_\alpha. \\
\end{cases}
\tag{6} \label{DeffStar}
$$
Because of \eqref{Disjoint}, $f^*$ is a well-defined function.
That $f^*$ has $p_n, n=0,1,\ldots$ as periods follows from \eqref{RightPeriod}.
We will prove that $f^*$ is continuous, using the $\epsilon-\delta$-definition.
Choose any $x \in D^*$, assume w.l.o.g that $x \in D_0$ (the proof works exactly analogous if $x \in D_\alpha$). We know that $x \notin D_\alpha$ from \eqref{Disjoint} and hence from Lemma 1 it follows that $x$ is also not an accumulation point of $D_\alpha$. That means there exists a $\delta(x) > 0$ such that $[x-\delta(x),x+\delta(x)] \cap D_\alpha = \emptyset$.
Now, given any $\epsilon > 0$, we choose the corresponding $\delta=\delta(x)$ for the continuity proof. For any $y \in D^*$ with $\lvert x-y \rvert < \delta$ we have $y \in D_0$, so $\lvert f(x)-f(y)\rvert = 0 < \epsilon$.
So to recap, $f^*$ is a non-constant continous function on $D^*$, with periods $p_n, n=0,1,\ldots$
Now the definition of the function $f$ that fullfills all the conditions asked by the OP and has no minimal period is
$$D=D^*\setminus \{1\}, f: D \rightarrow \mathbb R, f(x)=f^*(x) \; \forall x \in D. \tag{7} \label{Deff}$$
$f$ is $f^*$ restricted, so is still continuous. $f$ is obviously still non-constant. $f$ does not have period $p_0=1$, as $0 \in D,$ but $0+1 \notin D$.
$f$ still has periods $p_n, n=1,2,\ldots$. To prove this, let's fix some $n \ge 1, x \in D$, then by \eqref{Deff} $x \in D^*$, so by \eqref{DefDStar},\eqref{RightPeriod} $x+p_n \in D^*$. By \eqref{DefDs} each $y \in D_s$ fullfills $y \ge s$, so by \eqref{DefDStar} and \eqref{DefAlpha} so we have $x \ge 0$. Since $p_n > 1$ by its definition, we have $x+p_n > 1$. That means $x+p_n \in D$, and $f(x)=f(x+p_n)$ follows from the peridocity of $f^*$.
So $f$ has a set of periods that have $1$ as infimum, but $1$ itself is not a period.
Could $f$ have a smaller period $p$ with $0 <p < 1$? Since $0 \in D$, that would imply $1 > 0+p \in D$. As can be seen from \eqref{DefDs} and $p_n \ge 1, n=0,1,\ldots$, the only element of $D_0$ that is smaller than $1$ is $0$, and the only element of $D_\alpha$ that is smaller than $1$ is $\alpha$ (because of \eqref{DefAlpha}). That would imply $p=\alpha$. But then we would have $2\alpha=\alpha+\alpha=\alpha+p \in D$, but from \eqref{DefAlpha} we have $\alpha < 2\alpha < 1$ and we've just seen that besides $0$ and $\alpha$ there are are no more elements in $D$ that are smaller than 1.
So we conclude that $f$ cannot have a period $p < 1$. That concludes the proof that the $f$ defined by \eqref{Deff} is non-constant, continuous and has a set of periods greater than $1$ that have $1$ as infimum, but no period $0 < p \le 1$ exists.
--------------------------------
What remains to be done is prove Lemma 1!
All the $D_s$ are just translates of $D_0$, so they have the same topoligical properties, so it is enough to show that $D_0$ is closed.
In the sum $\sum_{n=0}^{\infty}k_np_n$ from \eqref{DefDs} only finitely many $k_n$ are non-zero. We define for each integer $m \ge 0$
$$A_m=\{x:\in \mathbb R: x=\sum_{n=0}^{\infty}k_np_n, k_n \in \mathbb Z, k_n \ge 0, n=0,1,\ldots, \sum_{n=0}^{\infty}k_n = m\} \tag{8} \label{DefAm}$$
and we have
$$ \bigcup_{m=0}^\infty A_m=D_0. \tag{9} \label{UnionAm}$$
It's easy to see that for example
$$A_0=\{0\}, A_1=\{p_0,p_1,\ldots,p_n,\ldots\} \tag{10} \label{Amsmall},$$
each $A_m$ consists of the sum of exactly $m$, not necessarily distinct, $p_n, n=0,1,\ldots$.
So $p_1+p_1+p_5 = 2p_1+p_5 \in A_3$, for example.
If we define the natural numbers $\mathbb N:=\{z \in \mathbb Z, z \ge 0\}$ as the non-negative integers, then $\mathbb N^m$ is the set of $m$-tuples of non-negative integers.
With $\bar{v}=(v_1,v_2,\ldots,v_m) \in \mathbb N^m$ we can define
$$R(\bar{v})=\sum_{k=1}^mp_{v_k} \in A_m. \tag{11} \label{DefR}$$
Conversely, for each $x \in A_m$ there is a $\bar{v} \in \mathbb N^m$ with $x=R(\bar{v})$ as $x$ is the sum of exactly $m$ $p_n's$, and the $\bar{v}$ just can take the $m$ indices '$n$'.
As a last step, we restrict our choice of $\bar{v}$ from now on to
$$O_m = \{(v_1,\ldots,v_m) \in \mathbb N^m: v_1 \le v_2 \le \ldots \le v_{m-1} \le v_m\}, \tag{12} \label{DefOm}$$
the set of increasingly ordered tuples. Due to commutativity of addition, we get from \eqref{DefR} and what's been said afterwards that
$$A_m=\{R(\bar{v}): \bar{v} \in O_m\}. \tag{13} \label{TupleEquiv}$$
Note that the representation of each element in $A_m$ in \eqref{TupleEquiv} is not unique, usually many different $\bar{v} \in O_m$ can have the same $R(\bar{v})$ and thus represent the same element of $A_m$. What's important is that such a representation exists at all.
We now prove that each $A_m$ ($m \ge 0$) is closed. That's done by mathematical induction on $m$.
The statement is obviously true for $A_0$ (see \eqref{Amsmall}). By definition, the only accumulation point of $A_1$ (also \eqref{Amsmall}) is $p_0=1$, which is part of $A_1$, so $A_1$ is also closed. This serves as start of the induction. Let's assume $m \ge 2$ and that we have proved that $A_{m-1}$ is closed as the induction hypothesis.
We are looking for an accumulation point $x_a$ of $A_m$. That means there exists a sequence $x_1,x_2,\ldots,x_n,\ldots \in A_m$ with $\lim_{n \to \infty}x_n=x_a$.
By \eqref{TupleEquiv} we have $m$-tuples $\bar{v_1},\bar{v_2},\ldots \in O_m$ with $ \forall n \ge 0: R(\bar{v_n})=x_n$ and
$$\lim_{n\to\infty}R(\bar{v_n})=x_a. \tag{14} \label{LimR}$$
For clarity, for $k=1,2,\ldots,m$ the $k$-the component of the $m$-tuple $\bar{v_n}$ will be denoted by $v_n^{(k)}$. No exponentiation is being used in this solution, so this shouldn't be too confusing.
We consider especially the sequence $\{v_n^{(m)}\}_{n \ge 1}$. There are 2 cases:
$\lim_{n\to\infty}v_n^{(m)} = +\infty$, or
$\exists N \in \mathbb N: v_n^{(m)} = N \text{ for infinitely many } n$.
It's easy to see the conditions on those 2 cases are mutually exclusive but cover every possible $\{v_n^{(m)}\}_{n \ge 1}$.
Case 1:
We define $(m-1)$-tuples $u_n$ for each $n \ge 1$ by simply omitting the last component from $v_n$:
$$u_n^{(k)}=v_n^{(k)}, n=1,2,\ldots; k=1,2,\ldots,m-1 \tag{15} \label{Reduction}.$$
From \eqref{DefR} we get
$$R(\bar{v_n})=R(\bar{u_n})+p_{v_n^{(m)}}\; \forall n \ge 1 \tag{16} \label{InductionCase1}$$
and now we know that $\lim_{n\to\infty}R(\bar{v_n})$ exists (equation \eqref{LimR}) and because we are in case 1 we also know that $\lim_{n\to\infty}p_{v_n^{(m)}}$ exists (it equals $\lim_{n\to\infty}p_n = 1$), so we know that
$$\lim_{n\to\infty}R(\bar{u_n}) = \lim_{n\to\infty}\left(R(\bar{v_n}) - p_{v_n^{(m)}}\right) = \lim_{n\to\infty}R(\bar{v_n}) - \lim_{n\to\infty}p_{v_n^{(m)}}=x_a-1 \tag{17} \label{ReductionCase1}$$
exists and what value it has.
But by the definition of $u_n$ in \eqref{Reduction}, we have $u_n \in O_{m-1}$, so $y_n:=R(\bar{u_n}) \in A_{m-1}$ by \eqref{TupleEquiv}, applied to $m-1$ instead of $m$.
So we get $\lim_{n\to\infty}y_n$ exists, and it's either a point of $A_{m-1}$ itself (if the $y_n$ become constant after some time) or an accumulation point of $A_{m-1}$. But the induction hypothesis says that $A_{m-1}$ is closed, so in either case we have
$$\lim_{n\to\infty}y_n =x_a-1 \in A_{m-1}$$
So $x_a-1$ is a sum on exactly $m-1$ (not necessarily distinct) $p_n's, n=0,1,\ldots$. If we add $p_0=1$ to that we get $x_a$, which proves $x_a \in A_m$. That concludes case 1.
Case 2: ($\exists N \in \mathbb N: v_n^{(m)} = N \text{ for infinitely many } n$)
By the definition of $O_m$ in \eqref{DefOm} we see that $v_n^{(k)} \le N\, \forall k=1,2,\ldots,m$ and each $n$ with $v_n^{(m)} = N$. There is only a finite number of $\bar{v}$ with $v^{(k)} \le N\, \forall k=1,2,\ldots,m$, so there must be at least one such $\bar{v}$ that occurs infinitely often as $\bar{v_n}$. In other words, in case 2 the sequence $\{\bar{v_n}\}_{n \ge 1}$ has a constant subsequence. That means $\{R(\bar{v_n})\}_{n \ge 1}$ has a constant subsequence, so by \eqref{LimR} $x_a=R(\bar{v_n})$ for some $n$, in other words $x_a \in A_m$.
That concludes case 2, and also the induction step to prove that $A_m$ is also closed: an accumulation point $x_a$ of $A_m$ was in both cases found to be in $A_m$.
We are almost done now. Let $x$ now be an accumlation point of $D_0$, set $N=\lfloor x \rfloor +1$. Choose $\delta=N-x > 0$. Since $x$ is an accumulation point of $D_0$, there must be a sequence of points $x_1,x_2,\ldots,x_n,\ldots \in D_0 \cap (x-\delta, x+\delta)$ that converges to $x$.
But for $m \ge N$ we know from \eqref{DefAm} that each element of $A_m$ is the sum of at least $N$ values ($p_n$'s) that are all at least $1$, so each element of those $A_m$'s is at least $N$, so cannot lie in $(x-\delta, x+\delta)=(x-\delta, N)$.
By \eqref{UnionAm} that means all elements of the above mentioned sequence $x_1,x_2,\ldots,x_n,\ldots \in D_0 \cap (x-\delta, x+\delta)$ must come from $\bigcup_{m=1}^{N-1}A_m$.
This is a union of a finite number of $A_m$'s, so there must be an infinite subsequence $x_{s_n}$ coming from a single $A_m$. That subsequence of course also converges to $x$, but as we have proved before, that $A_m$ is closed, so we have $x \in A_m \subseteq D_0$.
That concludes the proof of Lemma 1 and the solution to the problem.
Best Answer
Your proof seems correct, but there is a much simpler way to prove the statement.
Fix $x \in \mathbb{R}$, and let $\varepsilon > 0$.
Because $f$ is continuous at $\zeta$, there exists $\eta > 0$ such that for all $y \in [\zeta-\eta, \zeta + \eta]$, $|f(y)-f(\zeta)| \leq \varepsilon$. Let $T$ be a period of $f$ such that $0< T < 2\eta$. There exists $N \in \mathbb{Z}$ such that $x + NT \in [\zeta-\eta, \zeta + \eta]$, so you deduce that $|f(x)-f(\zeta)| = |f(x+NT)-f(\zeta)| \leq \varepsilon$. Because this has to be true for all $\varepsilon > 0$, you deduce that $f(x)=f(\zeta)$.
Hence $f$ is constant.