If $exists$ a surjection from nonempty set $A$ onto $mathcal{P}(mathbb{Z_+})$, then $A$ is uncountable

I wanted to show this to use it in some other proofs. I believe it holds. My questions are, if this proposition is true and if the proof sufficiently shows it holds.

Background

From a previous proof, it uses the result that $mathcal{P}(mathbb{Z_+})$, the power set of the positive integers, is uncountable.

Theorem 7.1 nonempty set A is countable $iffexists$ a surjective function $h: mathbb{Z_+} to A$ (citation: Munkres Topology Chapter 7)

Proof by Contradiction:

Let $A$ be a nonempty set and suppose $A$ is countable.

Assume $exists$ a surjective function $h: A to mathcal{P}(mathbb{Z_+})$. Since $A$ is countable, $exists$ a surjective function $f: mathbb{Z_+} to A$, by Theorem 7.1.

Consider the composition $hcirc f: mathbb{Z_+} to mathcal{P}(mathbb{Z_+})$. Since $f$ and $h$ are surjections, it follows that $hcirc f$ is a surjection ($cin mathcal{P}(mathbb{Z_+}) Rightarrow exists ain A$, such that $c = h(a) Rightarrow exists nin mathbb{Z_+}$, such that $c = h(f(n))$.

This implies that $mathcal{P}(mathbb{Z_+})$ is countable, by Theorem 7.1, a contradiction. Hence, $A$ is uncountable.