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.