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Let $(X,\tau)$ be a $T_1$-space and $X$ is an infinite set. Then $(X,\tau)$ has a subspace homeomorphic to $(\mathbb{N},\tau_2)$, where $\tau_2$ is either the finite-closed topology or the discrete topology.
Update attempt: As suggested from Daniel Fischer's comment, a solution is presented in the answer section.
Suppose that (a countable space) $X$ contains no infinite cofinite subspace. We will show it has a countable discrete subspace.
So we start by finding a non-empty open subset $U_0$ of $X$ such that $X \setminus U_0$ is infinite, and we pick $x_0 \in U_0$.
Next we will choose $x_0, x_1,x_2, \ldots$ and open sets $U_0, U_1, U_2, \ldots$ by recursion such that when we have chosen $x_0, \ldots, x_{n-1}$ ,$U_0, \ldots U_{n-1}$ in such a way that for all $0 \le i,j \le n-1$:
$x_i \in U_i$.
$x_j \notin U_i$ for $j \neq i$.
$A_{n-1} = X\setminus \bigcup\{U_m: 0\le m \le n-1\}$ is infinite.
Then we note that $A_{n-1}$ does not have the cofinite topology, so it has a relatively open non-empty subset $O$ with infinite complement in $A_{n-1}$, and so by $T_1$-ness (we have to avoid finitely many points) we have $U_n$ open in $X$ with $A_{n-1} \setminus (X \cap U_n)$ infinite and $U_n \cap \{x_0,\ldots,x_{n-1}\} = \emptyset$. This defines $X_n$ and finally we pick $x_n \in U_n \cap A_{n-1}$.
The last condition is needed to keep the recursion going and the first two show that the set $Y := \{x_n: n \in \mathbb{N}\}$ is an infinite discrete subspace of $X$ (as $U_n \cap Y = \{x_n\}$ for all $n$ and so all singletons are open in $Y$).
