I want to prove or disprove $\min{S_1}=\min{S_2}\Longleftrightarrow\min{S_1}\in S_2\land\min{S_2}\in S_1$ but I don't know where to start.
Maybe take cases like $\min{S_1}<\min{S_2}$, $\min{S_1}>\min{S_2}$ and $\min{S_1}=\min{S_2}$ and see where it is true?
(1). If $\min S$ exists then for any $x$ we have $(x\in S\implies \min S\leq x).$ Hence with $x=\min S_1$ and $S=S_2$ we have $$\min S_1\in S_2\implies \min S_2\leq \min S_1.$$ Interchanging the subscripts we have $$\min S_2\in S_1\implies \min S_1\leq \min S_2.$$ $$\text {Therefore }\quad (\min S_1\in S_2\land \min S_2\in S_1)\implies$$ $$\implies (\min S_2\leq \min S_1\land \min S_1\leq \min S_2)\implies \min S_1=\min S_2.$$
(2). If $\min S_1=\min S_2$ then $(\min S_1=\min S_2\in S_2 \land \min S_2=\min S_1\in S_1).$
To show a proposition is true if and only if another one holds, you need to show it for both directions.
The forward direction is rather straightforward. If we assume that $minS_1 = minS_2$ then let's call it $k = min S_1 = min S_2$. Then by definition we know that $k \in S_1$ since it is the min of the set. Likewise, we know that $k \in S_2$ by the same rationale.
The other direction, namely that $min S_1 \in S_2$ and $min S_2 \in S_1$ is a little more difficult to show. Let's presume that $m_1 = min S_1 \neq min S_2 = m_2$. Then we must have that either $m_1 > m_2$ or vice versa.
Assume that $m_1 > m_2$: Since we have $m_1 > m_2$ and $m_1 = min S_1$ then it must be that $m_2 \not\in S_1$, for if $m_2 \in S_1$ then that would violate $m_1$ being the min of the set. However this is a contradiction, since we know that $m2 \in S_2$, so we must have that $m_1 <= m_2$.
If, however, we have $m_1 < m_2$ then we can see that $m_2 = min S_2$ so it must be that $m_1 \not\in S_2$ as otherwise we would have that $m_1 = min S_2$. But this of course also contradicts what we know to be true, namely that $m_1 \in S_2$. So we must have that $m_1 >= m_2$.
If $m2 <= m_1 <= m2$ we must have that $m_1 = m_2$. As neeeded.