How can I prove that a set of natural numbers always have a minimum? Let's say I have a finite not-empty set named A, which is a set of natural numbers.
How do I prove it has a minimum?
(In Calculus)

You can't. Unless you add the condition that $A\ne \emptyset$.
Assume $A\subseteq \mathbb N$ has no monimal element.
Then prove by induction that $$\{1,\ldots,n\}\cap A=\emptyset$$ holds for all $n\in \mathbb N$.
The induction stecp $n\to n+1$ goes as follows: $\{1,\ldots,n\}\cap A$. If $n+1\in A$, this would imply that $n+1$ is a minimal element of $A$, contrary to the assumption. Therefore $n+1\notin A$ and hence $\{1,\ldots,n+1\}\cap A=\emptyset$.
Then note that $\{1,\ldots,n\}\cap A$ for all $n$ implies $A=\emptyset$ (as $n\in A$ implies $n\in \{1,\ldots,n\}\cap A$).
