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(In the following proof a one-to-one correspondence means a bijection; and it says that because we found an injective function such that it is not surjective so there is not any other function such that it biject the two sets $X$ and $P(X)$. Like saying that the function $f: \mathbb{N} \to \mathbb{Q} $ defined by $f(n)=n$ is not bijective so never there is a one-to-one correspondence between the two sets $\mathbb{N}$ and $\mathbb{Q} $.

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Didn't I understand it correctly or am I right?

The excerpt is taken from the book Basic Set Theory by Nikolai Konstantinovich Vereshchagin and Alexander Shen, page 27.

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    The proof shows, by contradiction, that there is no bijection $X \rightarrow \mathcal{P}(X)$. Thus, given any function $X \rightarrow \mathcal{P}(X)$, it is not bijective, hence either not injective or not surjective. – 2017-02-01
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    You didn't understand it correctly. The proof starts with the assumption that $\varphi$ is a bijection, and deduces a contradiction. That shows that no bijection exists. Essentially, the argument shows that **no** map $\varphi \colon X \to P(X)$ is _surjective_. If you drop the assumption that $\varphi$ is a bijection from the start of the proof and replace it with "Let $\varphi \colon X \to P(X)$ an arbitrary map.", the argument shows that $Z \notin \varphi(X)$, so $\varphi$ is not surjective. Since $\varphi$ was arbitrary, no map is a surjection. – 2017-02-01
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    I think that it is good to include also the source of the problem. I have added it to your post - I hope that I have identified the book correctly. – 2017-02-01

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Your understanding is not correct. In fact, your example is false: there does exist a one-to-one correspondence between $\mathbb{N}$ and $\mathbb{Q}$. This is a little hard to see, but an easier example is taking $f:\mathbb{N}\to\mathbb{N}\cup\{-1\}$ given by $f(n)=n$. Then $f$ is injective but not surjective, but there does exist a one-to-one correspondence between $\mathbb{N}$ and $\mathbb{N}\cup\{-1\}$: namely, define $g:\mathbb{N}\to\mathbb{N}\cup\{-1\}$ by $g(n)=n-1$. So just having one injection that is not a surjection does not mean you can't have some different injection that is surjective as well.

I suggest you take some time to read the proof of Theorem 8 carefully. The proof is a proof by contradiction: it starts by assuming there exists some one-to-one correspondence $\varphi$, and then derives a contradiction (by proving that $\varphi$ cannot actually be surjective). Therefore the assumption that there exists a one-to-one correspondence must be false. Nowhere does this argument involve giving an example of an injection that is not surjective as you suggest; instead, it is an argument that applies to any one-to-one correspondence (if such a thing existed) and derives a contradiction.

(Incidentally it is not actually necessary to phrase this proof as a proof by contradiction--you can instead reorganize it so that it directly proves that no map $X\to P(X)$ is a surjection.)