Existence of a denumerble partition. Is the following Proof Correct?

NOTE: In the text prior to the following exercise we have already proved that $\mathbf{Z^+}\times\mathbf{Z^+}$ is equinumerous to $\mathbf{Z^+}$.

Theorem. Given that $A$ is denumerable. There exists a partition $P$ of $A$ such that $P$ is denumerable and for all $X\in P$, $X$ is denumerable.

Proof. Since $A$ is denumerable there exists a bijection $h:\mathbf{Z^+}\to A$ furthermore from our initial discussion in $7.1$ we know that $\mathbf{Z^+}\times\mathbf{Z^+}\thicksim\mathbf{Z^+}$ consequently there exists a bijection $f:\mathbf{Z^+}\times\mathbf{Z^+}\to\mathbf{Z^+}$ in summary we have a bijection $h\circ f:\mathbf{Z^+}\times\mathbf{Z^+}\to A$. Now consider the following definition.
$$\forall k\in\mathbf{Z^+}\left(\mathcal{E}_k = \{(h\circ f)(k,j)\in\mathbf{Z^+}|j\in\mathbf{Z^+}\}\right)$$

We now prove that $P = \cup_{k\in\mathbf{Z^+}}\mathcal{E}_k = A$. Evidently $P \subseteq A$. Let $a\in A$ be arbitrary, the surjectivity of $A$ implies that for some $(i,j)\in\mathbf{Z^+}\times\mathbf{Z^+}$, $(h\circ f)(i,j) = a$ equivalently $a\in \mathcal{E}_i$ implying that $A\subseteq P$ and by extension $P = A$.

We now prove that $P$ is pairwise-disjoint. Assume for some $i\in\mathbf{Z^+}$ and $j\in\mathbf{Z^+}$ that $\mathcal{E}_i\cap\mathcal{E}_j\neq\varnothing$ and $i\neq j$ equivalently there exists some $q\in\mathcal{E}_i\cap\mathcal{E}_j$ thus for some $m,n\in\mathbf{Z^+}$, $(h\circ f)(i,m)=q=(h\circ f)(j,n)$ but the injectivity of $h\circ f$ implies that $(i,m) = (j,n)$ and by extension $i = j$ resulting in a contradiction.

In addition given that $h\circ f$ is a bijection it is not difficult to see that for any $k\in\mathbf{Z^+}$, $g_k:\mathbf{Z^+}\to\mathcal{E}_k$ defined as follows
$$g_k(i) = (h\circ f)(k,i)$$
is a valid bijection from $\mathbf{Z^+}$ to $\mathcal{E}_k$ implying that $\mathcal{E_k}$ is denumerable.
Yes it is a proof .Why do you doubt it ?
Yes it is a proof .Why do you doubt it ?