compact and countably compact

As we know that a topological space $( X, \tau)$ is said compact if every open cover for $X$ has a finite subcover, and a topological space $( X, \tau)$ is said countably compact if every countably open cover for $X$ has a finite subcover. So, every compact space is countably compact.

Is there example to show that converse is not true?

In Seymour book, present a example to show that converse is not true.

He take $\tau$ a topology on natural number such that is generated by $ \{ 1,2 \}, \{3,4\}, \{5,6\}....$.

and said that $( ‎‎\mathbb{N}, \tau)$ is a countably compact space.

But, I think, every countably open cover has not a finite subcover

You are correct: that space is not countably compact. Indeed, a space that is countable and countably compact is automatically compact, since every open cover certainly has a countable subcover.

One of the simpler examples of a space that is countably compact but not compact is $\omega_1$, the space of countable ordinal numbers, with the order topology. If for each $\alpha<\omega_1$ we let $U_\alpha=[0,\alpha)$, then $\{U_\alpha:\alpha<\omega_1\}$ is an open cover of $\omega_1$ with no finite subcover. However, every infinite subset of $\omega_1$ has a limit point in $\omega_1$, so $\omega_1$ contains no infinite closed discrete set and is therefore countably compact.

To see this, suppose that $A\subseteq\omega_1$ is infinite. Let $\alpha_0=\min A$, and for each $n\in\omega$ let $\alpha_{n+1}=\min\{\alpha\in A:\alpha>\alpha_n\}$. Then $\langle\alpha_n:n\in\omega\rangle$ is a strictly increasing sequence in $A$. Let $\alpha=\sup_{n\in\omega}\alpha_n$; then $\alpha<\omega_1$, and $\alpha$ is easily seen to be a limit point of $A$.