In $\mathbb{N}$, we define the topology defined by $$T=\emptyset \cup\{\{0,1,2...,n\} :\space n\in \mathbb{N}\}\cup \mathbb{N}$$
Now I want to prove that $(\mathbb{N},T)$ is not compact.
Suppose $(\mathbb{N},T)$ is compact, then there exists a collecion of open sets $U_i$ so that $\mathbb{N}=\bigcup\limits_{i=1}^{n} U_{i}$. $U_i$ are of the form $\{0,1,2,3,...,i\}$ then there exists $U_j>U_i$ for all $i=1,...,n$ $i\neq j$. Now take $M\in \mathbb{N}$ so that $M>x$ for all $x\in U_j$ . This $M$ exists because $U_j$ is a finite subset of $\mathbb{N}$ and therefore is bounded. Then $M\in \mathbb{N}$ however $M\not\in\bigcup\limits_{i=1}^{\infty} U_{i}$. By contradiction I have proven that $(\mathbb{N},T)$ is not compact.
Is this proof correct? I was doubting if I could take $\mathbb{N}$ as a $U_i$
EDIT
Second try:
Take the open cover $\bigcup\limits_{i=1}^{\infty} U_{i}=\mathbb{N}$ where $U_i=\{1,2,...,i\}$ then by any $n\in\mathbb{N}$, the open subcover $\bigcup\limits_{i=1}^{n} U_{i}$ does not contain $\mathbb{N}$ and therefore $(\mathbb{N},T)$ is not compact.
No, it is not correct. Asserting that $\mathbb N$ is compact means that for every family $(U_\lambda)_{\lambda\in\Lambda}$ of open subsets of $\mathbb N$ whose union contains $\mathbb N$, there is a finite subset $F$ of $\Lambda$ such that $\mathbb{N}\subset\bigcup_{\lambda\in F}U_\lambda$. So, assertint that $\mathbb N$ is not compact means that there is a family $(U_\lambda)_{\lambda\in\Lambda}$ of open subsets of $\mathbb N$ whose union contains $\mathbb N$ for which there is no finite subset $F$ of $\Lambda$ such that $\mathbb{N}\subset\bigcup_{\lambda\in F}U_\lambda$. This is indeed true. Take $\Lambda=\mathbb N$ and, for each $n\in\Lambda$, $U_n=\{0,1,2,\ldots,n\}$.