Optimal stopping times $\tau_{\min}^{K}$ for $H_{t}^{K}=\frac{(K-S_{t})^{+}}{(1+r)^{t}}$

Question

$H^{K}$ is the process $H_{t}^{K}=\frac{(K-S_{t})^{+}}{(1+r)^{t}}$

Note that $S$ is a stochastic process, and $K,r,t$ are constants where $t\in \{1,...,T\}$ for some $T \in \mathbb N$ and $ r>0$

Further let $\tau_{\min}^{K}$ be the minimal optimal stopping time for $H^{K}$, i.e. $E[H_{\tau_{\min}^{K}}^{K}]$ maximizes over all stopping times

1. Let $L \geq K$: Show that $\tau_{\min}^{K}\geq \tau_{\min}^{L}$

2. Show that $\inf_{K \geq 0} \tau_{\min}^{K} =0$ $ P-$a.s.

3. Let $\mathcal{F}_{0}=\{\Omega,\varnothing\}$, use $2.$ that there exists $K_{0}\geq 0$ so that $ \tau_{\min}^{K} =0$ $P-$a.s. for all $K \geq K_{0}$

My idea:

$1.$ Note that $(K-S_{t})^{+}\leq (L-S_{t})^{+}$ and thus any maximizer of $(K-S_{t})^{+}$ would also maximize $(L-S_{t})^{+}$, hence $\tau_{\min}^{K}$ is an optimal stopping time for $H^{L}$. Since $\tau_{\min}^{L}$ is the minimal optimal stopping time for $H^{L}$, it follows $\tau_{\min}^{K}\geq \tau_{\min}^{L}$. Is my reasoning sound?

$2.$ Not sure if I could use a contradiction?

$3.$ Since $\mathcal{F}_{0}=\{\Omega,\varnothing\}$ and $\inf_{K \geq 0} \tau_{\min}^{K} =0$ $ P-$a.s., so that there exists a constant stop time, it follows that for some $K_{0}$ that $\tau_{\min}^{K_{0}}=0$ and by question $1.$ it follows $\tau_{\min}^{K}=0$ for all $K \geq K_{0}$

I am looking for an answer for $2.$ and critique/correction of my proofs for $1.$ and $3.$