Definition: sD means $s$ is a digit, i.e. $s$ is one of $0,1,\dots,9$. k Nat means $k\in\bf N$.
Proposition: If sD <-> k Nat then sD and k Nat.
I don't really understand what I have to prove. Do I have to prove that ¬SD and ¬k Nat isn't possible? The hint I was given is to use induction but I don't see how.
With the information that you provided in your question, from a logical point of view the implication mentioned in your question is false.
What you could prove is that
(sD⇔k Nat)⇒((sD∧k Nat)∨(¬(sD)∧¬(k Nat)))
That is
¬(sD⇔k Nat)∨(sD∧k Nat)∨(¬(sD)∧¬(k Nat))
¬((sD⇒k Nat)∧(k Nat⇒sD))∨(sD∧k Nat)∨(¬(sD)∧¬(k Nat))
¬((¬(sD)∨(k Nat))∧(¬(k Nat)∨(sD)))∨ (sD∧k Nat)∨(¬(sD)∧¬(k Nat))
(sD∧¬(k Nat))∨(k Nat ∧¬(sD))∨(sD∧k Nat)∨(¬(sD)∧¬(k Nat))
The last statement is true because s could be a digit or not, k could be a natural number or not, and the statement above covers all 4 possible combinations.