Subspace of a topological space with cofinite topology I'm thinking about topological spaces with a cofinite topology $S$ and I proved a subspace ($M$) of $S$ with a relative topology has a discrete topology if $M$ is finite. Then what if $M$ is infinite?

I tried $\mathbb{Z}$ with a cofinite topology and its subset $\mathbb{N}$ and I think $\mathbb{N}$ has a cofinite topology as its relative topology too. Is this correct?
You are correct.  The open sets of $\mathbb{Z}$ under the cofinite topology are the $U$ such that $\mathbb{Z} \setminus U$ is finite.  The open sets of $\mathbb{N}$ under the subspace topology are of the form $\mathbb{N} \cap U$ for some open subset $U$ of $\mathbb{Z}$; so will $\mathbb{N} \setminus (\mathbb{N} \cap U)$ be finite?  Of course.  If it were infinite, then $U$ would not have finite complement in $\mathbb{Z}$, a contradiction!  

This means every open set in $\mathbb{N}$ w.r.t. the subspace topology is cofinite in $\mathbb{N}$.  Now we need to establish that every cofinite subset of $\mathbb{N}$ appears as an open set in the subspace topology:

Suppose a subset $V$ is cofinite in $\mathbb{N}$.  This corresponds to a cofinite subset of $\mathbb{Z}$ that intersects with $\mathbb{N}$ to give $V$, namely $V \cup \{x \in \mathbb{Z} \ | \ x \leq 0 \}$.  So $V$ is open in $\mathbb{N}$ under the subspace topology.

We have thus established equality between the sets of open sets under the two topologies.
A subspace of cofinite space is cofinite. Let $(X,T)$ be cofinite and $(Y,S)$ be subspace. $S$ is cofinite if $Y \cap G$ has finite complement in $Y$. Complement of ($Y \cap G$) in $Y = Y \cap (\text{complement of }G)$, which is finite. Thus $(Y,S)$ is a cofinite space. [Note that complement of $G$ is finite].