Irregular Induction Theorem for $\mathbb{N}\times\mathbb{N}$

Question

I am trying to prove this irregular induction theorem that would help prove a recursion theorem I am working on.

Can you help? Here is the theorem:

$\forall X (\forall x \in \mathbb{N} (\langle x,0 \rangle \in X \land \forall y \in \mathbb{N} (\langle x,S(y) \rangle \in X)) \rightarrow \mathbb{N} \times \mathbb{N} \subseteq X)$

I don't know if it is proveable but it seems intuitively correct. Using the second-order I think I can prove it but I want to try first order first.

Here is my work before I got stuck:

1. $\forall x \in \mathbb{N} (\langle x,0 \rangle \in X \land \forall y \in \mathbb{N} (\langle x,S(y) \rangle \in X))$
   Premise

2. $\forall x (x \in \mathbb{N} \rightarrow (\langle x,0 \rangle \in X \land \forall y \in \mathbb{N} (\langle x,S(y) \rangle \in X$))
   Notation (1)

3. $x \in \mathbb{N}$
   Premise

4. $x \in \mathbb{N} \rightarrow (\langle x,0 \rangle \in X \land \forall y \in \mathbb{N} (\langle x,S(y) \rangle \in X$)
   Universal Instantiation $x$/$x$ (2)

5. $\langle x,0 \rangle \in X \land \forall y \in \mathbb{N} \langle x,S(y) \rangle \in X$
   Modus Ponendo Ponens (4)(3)

6. $\forall y \in \mathbb{N} \langle x,S(y) \rangle \in X$
   Simplification (5)

7. $\forall y (y \in \mathbb{N} \rightarrow \langle x,S(y) \rangle \in X)$
   Notation (6)

8. $y \in \mathbb{N}$
   Premise

9. $y \in \mathbb{N} \rightarrow \langle x,S(y) \rangle \in X$
   Universal Instantiation $y$/$y$ (8)

10. $\langle x,S(y) \rangle \in X$
    Modus Ponendo Ponens (9)(8)

I can only deduce that $\mathbb{N}\times\mathbb{N}\subset\text{dom}(X)\times\text{im}(X)$ but since $\text{dom}(X)\times\text{im}(X)$ is not necessarily a subset of $X$ it lead nowhere.