Prove that identity map f from (X,T) (topological space) to (X,T') (cofinite topological space) is continuous if (X,T) is a T1 space So following is the proof that I have come up with but I'm not sure if it's correct or not. I'd be really grateful if you could validate/invalidate it.
To prove: Given $(X,T)$ is $T_1$, prove that $f$ is continuous
Proof: We know that Cofinite topologies are $T_1.$
So, $(X, T')$ is  $T_1.$
Let us assume that $f$ is not continuous, so there exists $y \subset X$  which is open in $T'$ s.t. $f^{-1}(y),$ which would be y itself since $f$ is an identity map, is not open in $T.$
Any open set in cofinite topology is of the form $X-Z$ where $ Z$ is finite. Let $X-Z$ be open in $(X,T')$. It's preimage in $(X,T)$ would also be $X-Z$ (since $f$ is identity map). Let this set $ X-Z$ be set $y$ which is open in $T'$ but not in $T$.
Given that $(X,T)$ is $T1$, so by definition all finite sets are closed. Since $Z$ is finite, it is closed. By the definition of closed sets, complement of $Z$ should be open. So, $X-Z$ is open in $T$.
But this is a contradiction since we had assumed $X-Z$ to not be open in $T.$ 
So, our assumption was wrong. Hence, proved?
 A: Your argument is stated rather clumsily and has one very small oversight, but it’s basically correct. The oversight is when you say that any set open in the cofinite topology is of the form $X\setminus Z$ for some finite $Z$: this is true for every non-empty open set in the cofinite topology. Of course the empty set won’t be a problem, so the oversight doesn’t really harm the proof.
Here’s one way to write up the same argument a little more clearly.

Observe that since $f^{-1}[U]=U$ for every $U\in T'$, $f$ is continuous if and only if $T'\subseteq T$. Suppose not; then there is a $U\in T'\setminus T$. Clearly $U\ne\varnothing$, so $U=X\setminus F$ for some finite $F\subseteq X$. But $T$ is $T_1$, and finite sets in $T_1$ spaces are closed, so $F$ is closed in $\langle X,T\rangle$, and therefore $U=X\setminus F\in T$. This contradiction shows that $U\in T$ and hence that $f$ is continuous.

In fact, if you examine the argument closely, you can see that we didn’t really need contradiction.

Observe that since $f^{-1}[U]=U$ for every $U\in T'$, $f$ is continuous if and only if $T'\subseteq T$. Let $U\in T'$. Then either $U=\varnothing$, in which case certainly $U\in T$, or $U=X\setminus F$ for some finite $F\subseteq X$. But $T$ is $T_1$, and finite sets in $T_1$ spaces are closed, so $F$ is closed in $\langle X,T\rangle$, and therefore $U=X\setminus F\in T$. Thus, $T'\subseteq T$, and $f$ is continuous.

