Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$.

I only really see one way to solve this problem, though how you write the details can vary considerably. You asked for an "elegant" solution, so here's my arrangement.

Key Lemma: Let $y \in (0, 1)$ and $k \in \mathbb{Z}_{\geq 2}$. Then $$\{yk^n \pmod{1} : n \in \mathbb{Z}_{\geq 0}\} \cap [1/k^2, 1-1/k^2] \neq \varnothing.$$

Proof: Consider the base $k$ expansion of $y$, so $y = 0{.}y_1 y_2 y_3{\ldots}$ and $yk^n \pmod{1} = 0{.}y_{n+1} y_{n+2}{\ldots}$. If the latter expression is not in $[1/k, 1-1/k]$, we must have $y_{n+1} = 0$ or $y_{n+1} = k-1$. So, we may suppose that $y_{n+1} \in \{0, k-1\}$ for all $n \in \mathbb{Z}_{\geq 0}$. Since $y \neq 0, 1$ by assumption, we have some $N$ where $y_{N+1} \neq y_{N+2}$. Hence $yk^N \pmod{1} = 0{.}y_{N+1} y_{N+2}{\ldots} = 0{.}0(k-1){\ldots}$ or $0{.}(k-1)0{\ldots}$. In the first case, $yk^N \pmod{1} \in [1/k^2, 1/k]$, while in the second case, $yk^N \pmod{1} \in [1-1/k, 1-1/k^2]$. $\Box$

Now set $\delta = \sin(\pi/k^2) > 0$. For all $x \in (0, \pi)$, by the key lemma applied to $y = x/\pi$, we have some $n \in \mathbb{Z}_{\geq 0}$ such that $xk^n \pmod{\pi} \in [\pi/k^2, \pi-\pi/k^2]$. From the unimodality and symmetry of $\sin$, it follows that $|\sin(xk^n)| \geq |\sin(\pi/k^2)| = \delta$.

First solution of the problem

We prove that for each $x \in(0,\pi)$, there exists $n,y\in \mathbb{N}$ such that $$\pi y + \frac{\pi}{k^2} \le x k^n\le \pi y +\pi - \frac{\pi}{k^2} \iff y +\frac{1}{k^2}\le k^n\frac{x}{\pi} \le y+1-\frac{1}{k^2} \tag{1}$$ because $$(1) \iff |\sin(xk^n)| \ge \sin \left(\frac{\pi}{k^2} \right)$$

We write $\frac{x}{\pi}$ in base-k numeral system (aka: radix system) as follows $$\frac{x}{\pi} =\overline{0,s_1s_2s_3...s_n...}^{(\mathbf{k})} = \sum_{i=1}^{+\infty}\frac{s_i}{k^i}=\frac{s_1}{k}+\frac{s_2}{k^2}+\frac{s_3}{k^3}+...+\frac{s_n}{k^n}+...$$ and then $$k^n\frac{x}{\pi} =\overline{s_1s_2s_3...s_n,s_{n+1}s_{n+2}...}^{(\mathbf{k})}$$ The integer part $\left[k^n\frac{x}{\pi}\right]$ and the fractional part $\{k^n\frac{x}{\pi}\}$ of $k^n\frac{x}{\pi}$ are defined as $$a_n =\left[k^n\frac{x}{\pi}\right] = \overline{s_1s_2s_3...s_n}^{(\mathbf{k})}$$ $$\epsilon_n=\left\{k^n\frac{x}{\pi}\right\} = \overline{0,s_{n+1}s_{n+2}...}^{(\mathbf{k})}$$

We notice that $0 \le \epsilon_n < 1$ and $k^n\frac{x}{\pi} = a_n + \epsilon_n $.

Case 1: $\frac{x}{\pi}$ is a terminating decimal in the base-k. Suppose $\frac{x}{\pi}$ has $N$ numbers after the $0$. Then, $\frac{x}{\pi}$ can be written as

$$\frac{x}{\pi} =\overline{0,s_1s_2s_3...s_N}^{(\mathbf{k})} = \frac{s_1}{k}+\frac{s_2}{k^2}+\frac{s_3}{k^3}+...+\frac{s_N}{k^N}$$ with $ 1 \le s_N \le (k-1)$.

Choose $y = a_{N-1}$ then $\epsilon_{N-1} = \overline{0,s_N}^{(\mathbf{k})} = \frac{s_N}{k}$. We can verify easily that $(y,n) = (a_{N-1},N-1)$ satisfy (1). Indeed, we have

$$a_{N-1}+\frac{1}{k^2} \le a_{N-1}+\frac{1}{k} \le a_{N-1}+ \frac{s_{N-1}}{k} = a_{N-1}+ \epsilon_{N-1} = k^{N-1} \frac{x}{\pi}$$ and $$k^{N-1} \frac{x}{\pi} = a_{N-1}+ \epsilon_{N-1} = a_{N-1}+ \frac{s_{N-1}}{k} \le a_{N-1}+ \frac{k-1}{k} \le a_{N-1}+ 1- \frac{1}{k^2}$$

Case 2-a: $\frac{x}{\pi}$ is a non-terminating decimal (aka repeating decimal) in the base-k and there exists an $n$ such that $1 \le s_{n+1} \le (k-2)$.

Then, $\epsilon_n = \overline{0,s_{n+1}s_{n+2}...}^{(\mathbf{k})}$ satisfies $\frac{1}{k} \le \epsilon_n \le \frac{k-1}{k}$. Choose $y=a_n$, we have

$$a_n +\frac{1}{k} \le k^n\frac{x}{k\pi} \le a_n +\frac{k-1}{k} $$ $$\iff a_n +\frac{1}{k} \le a_n+ \epsilon_n \le a_n +\frac{k-1}{k} \tag{2}$$

From (2), we can deduce that (1) holds true. Indeed, we have

$$a_n+\frac{1}{k^2 } \le a_n+\frac{1}{k} $$ and $$a_N+ \frac{k-1}{k} \le a_N+ 1- \frac{1}{k^2}$$

Case 2-b: $\frac{x}{\pi}$ is a non-terminating decimal (aka repeating decimal) in the base-k and there doesn't exist an $n$ such that $1 \le s_{n+1} \le (k-2)$.

So for all $i=1,...,+\infty$ we have $s_i = 0$ or $s_i=(k-1)$

Let $u$ is the first value such that $s_u = 0$ (if $u$ doesn't existe, then $s_u = (k-1)$ for all $u$. As we suppose $\frac{x}{\pi}$ is a non-terminating decimal, we have $\frac{x}{\pi} =\sum_{i=1}^{+\infty}\frac{(k-1)}{k^i} \rightarrow 1$ which is a contradiction.).

Let $v$ is the first value such that $v>u$ and $s_v = (k-1)$. (If $v$ doesn't existe, then $s_v = 0$ for all $v>u$. Then $\frac{x}{\pi}$ is not a non-terminating decimal $\implies$ contradiction). Hence, $\frac{x}{\pi}$ can be written as

$$\frac{x}{\pi} = \overline{0,s_{1}s_{2}...s_{u-1}00...00(k-1)s_{v+1}....}^{(\mathbf{k})}$$ In particular, $$k^{v-1}\frac{x}{\pi} = \overline{..s_{v-2},0(k-1)s_{v+1}..}^{(\mathbf{k})}$$

Let's $y = a_{v-1}$ and $n = v-1$, we have $$k^n\frac{x}{\pi} = a_{v-1} + \overline{0,0(k-1)s_{v+1}..}^{(\mathbf{k})}$$ and $$\overline{0,0(k-1)s_{v+1}..}^{(\mathbf{k})} = \frac{k-1}{k^2} + ... \in \left( \frac{1}{k^2}, 1-\frac{1}{k^2} \right)$$ $$\implies a_n + \frac{1}{k^2} \le k^n\frac{x}{\pi} \le a_n + 1-\frac{1}{k^2}$$ So, (1) holds true.

Conclusion Hence, we have for all $x\in (0,\pi), \exists n\in \mathbb{N}$ such that $|\sin(xk^n)|\ge \sin(\frac{\pi}{k^2})$.

Note: We can prove $ \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\cos(xk^n)|\ge \cos \left( \frac{\pi}{k} \right)$ by using the same method. In fact, for the $|\cos(xk^n)|$, the proof is less difficult, it suffices to prove there exists $y,n\in\mathbb{N}$ such that $y \le k^n\frac{x}{\pi} \le y+1-\frac{1}{k}$.

Hence, we have that $\forall x \in ]0,\pi[, \exists n,m \in \Bbb N$ such that $$|\sin(xk^n)|\ge \sin\left( \frac{\pi}{k^2} \right)$$ $$|\sin(xk^m)|\le \sin\left( \frac{\pi}{k} \right)$$

Not a proof of the statement but close.
Also, one should notice that the statement does not hold water for numbers in the form $x=\frac{\pi*m}{k}$ where $m <=k$
The question remains if these are the only "bad" x's.
For the proof, let $k_n$ be an increasing sequence of integers and $\sigma < \frac{1}{\sqrt2}$.
Let $E = \{x; \overline {\lim}_{n \to \infty} |\sin(k_{n}x)| < \sigma \}$.
Lastly, let $\chi_{E}$ be characteristic function of $E$.
Then $m(E)=0$.
To see this, note that on one hand $$\int_{0}^{2\pi}\chi_{E} \cos(2k_{n}x) dx$$ tends to $0$ because this is just a Fourier coefficient of integrable function.
At the same time since $\cos(2k_{n}x) = \frac { 1-2\sin^{2}k_nx}{2}$ one gets: $$\int_{0}^{2\pi}\chi_{E} \cos(2k_{n}x) dx = \frac{1}{2}m(E) - \int_{0}^{2\pi}\chi_{E} \sin^{2}(k_{n}x)dx $$, so regrouping and taking upper limits: $$\frac{1}{2}m(E) <= \overline {\lim}_{n \to \infty}\int_{0}^{2\pi}\chi_{E} \cos(2k_{n}x) dx + \overline {\lim}_{n \to \infty}\int_{0}^{2\pi}\chi_{E} \sin^{2}(k_{n}x) dx $$ As remarked above the first term tends to $0$.
The second one can be estimated by passing with upper limit inside the integral (Fatou's Lemma) and using definition of $E$, to get $<= \sigma^{2}m(E)$.

//////////////////////

This comment relates to the reformulated version of the problem added by Michelle on 2/15/2021. The new statement suggest much simpler solution. One looks at unit vectors (complex numbers) on the unit circle and the angle between them. If two vectors have angle between them (mod $2\pi$) , say $\alpha$, which is less then $\frac{\pi}{k}$ then their image under $f$ maps this angle into $k*\alpha$ (mod $2\pi$). Hence, for any two vectors $z_1, z_2$ if $z_1 != z_2$ there is an $n$ such that the angle between $z_1^n$ and $z_2^n$ is at least $\frac{\pi}{k}$. This gives proper estimates on $\sigma$ that is $\sigma >= |1-e^{i*\frac{\pi}{4}}|$