Topology induced from Kuratowski Closure properties 
Proof:- From (c) $X\in \mathscr T$. $X\subset cl(X)[\because by (a)]$ and also we have $cl(X)\subset X$. Hence $cl(X)=X$ and $\emptyset \in \mathscr T$. 
Let $\{U_i\}_{i=1}^n$ be a collection of open sets in $\mathscr T$. For each $i$ there exists $C_i$:$cl(C_i)=C_i$ and $U_i=X\setminus C_i$ then $\bigcap_{i=1}^{n}U_i=X\setminus \bigcup_{i=1}^n C_i$. We have $cl(C_i)=C_i,i=1,2,3,..,n$. By (d), we have $\bigcap_{i=1}^{n}U_i \in \mathscr T$. How do I prove arbitrary union axiom? If there is an axiom $\bigcap_{\alpha \in \Lambda}cl(A_\alpha)=cl(\bigcap_{\alpha \in \Lambda}A_\alpha)$. I could prove. Can you please help me?
 A: We just have to show that $$\mathcal{C} = \{C \subseteq X: \operatorname{cl}(C) = C\}$$ obeys the axioms for closed sets. So we indeed (among other things) have to show that it's closed under arbitary intersections, as you're doing.
So suppose that $C_i, i \in I$ obeys $\operatorname{cl}(C_i) = C_i$ for all $i$.
Consider $C= \bigcap_{i \in I} C_i$. As for all $i$, $C \subseteq C_i$ we have for all $i$ (see lemma below) : $\operatorname{cl}(C) \subseteq \operatorname{cl}(C_i) = C_i$, so that taking intersections over $I$ again: $\operatorname{cl}(C) \subseteq \bigcap_{i\in I} C_i  =C$ and $C \subseteq \operatorname{cl}(C)$ is true by (a). So $\operatorname{cl}(C) = C$ and you're done.

Lemma: if $\operatorname{cl}$ is a Kuratowski closure operator and $A \subseteq B$ then $\operatorname{cl}(A) \subseteq \operatorname{cl}(B)$ ($\operatorname{cl}$ is monotone).

Proof: $A \subseteq B$ iff $A \cup B = B$ which implies $\operatorname{cl}(A \cup B) = \operatorname{cl}(B)$ and applying (d), $\operatorname{cl}(A) \cup \operatorname{cl}(B) = \operatorname{cl}(B)$ which implies $\operatorname{cl}(A) \subseteq \operatorname{cl}(B)$ again. Done.  
Finally note that $\mathcal{C}$ is closed under finite unions by (d) and $\emptyset \in \mathcal{C}$ is just axiom (c),  $X \in \mathcal{C}$ follows from (a).  
Now $\mathcal{T}= \{C^\complement \subseteq X: C \in \mathcal{C}\}$, as your text defines is closed under all unions and finite intersections by de Morgan's laws:
Suppose $O_i, i \in I$ are in $\mathcal{T}$ and so $O_i = C_i^\complement$ for $C_i \in \mathcal{C}$, and we know $\bigcap_i C_i \in \mathcal{C}$ too. Hence $$\bigcup_i O_i = \bigcup_i C_i^\complement = \left(\bigcap_i C_i\right)^\complement \in \mathcal{T}$$ as well. The finite union case is similar.
