In the case of compactness it makes no difference whether you use the topology of $X$ or the relative topology on the subset.
Proposition. Let $\langle X,\tau\rangle$ be any space, let $K\subseteq X$, and let $\tau_K$ be the relative topology on $K$; then $K$ is compact with respect to $\tau$ iff it is compact with respect to $\tau_K$.
Proof. Suppose first that $K$ is compact with respect to $\tau$, and let $\mathscr{U}\subseteq\tau'$ be a $\tau'$-open cover of $K$. For each $U\in\mathscr{U}$ there is a $V_U\in\tau$ such that $U=K\cap V_U$. Let $\mathscr{V}=\{V_U:U\in\mathscr{U}\}$; clearly $\mathscr{V}$ is a $\tau$-open cover of $K$, so it has a finite subcover $\{V_{U_1},\ldots,V_{U_n}\}$. Let $\mathscr{F}=\{U_1,\ldots,U_n\}$; $\mathscr{F}$ is a finite subset of $\mathscr{U}$, and
$$\bigcup\mathscr{F}=\bigcup_{k=1}^nU_k=\bigcup_{k=1}^n(K\cap V_{U_k})=K\cap\bigcup_{k=1}^nU_k=K\;,$$
so $\mathscr{F}$ covers $K$. Thus, $K$ is compact with respect to $\tau'$.
Now suppose that $K$ is compact with respect to $\tau'$, and let $\mathscr{U}\subseteq\tau$ be a $\tau$-open cover of $K$. For each $U\in\mathscr{U}$ let $V_U=K\cap U$, and let $\mathscr{V}=\{V_U:U\in\mathscr{U}\}$. $\mathscr{V}$ is a $\tau'$-open cover of $K$, so it has a finite subcover $\{V_{U_1},\ldots,V_{U_n}\}$. Let $\mathscr{F}=\{U_1,\ldots,U_n\}$; $\mathscr{F}$ is a finite subset of $\mathscr{U}$, and
$$\bigcup\mathscr{F}=\bigcup_{k=1}^nU_k\supseteq\bigcup_{k=1}^n(K\cap U_k)=\bigcup_{k=1}^nV_{U_k}=K\;,$$
so $\mathscr{F}$ covers $K$. Thus, $K$ is compact with respect to $\tau$. $\dashv$
Best Answer
If $X$ is infinite, let $x \in X$ ; it follows that since $\{x\}$ is finite, $X \backslash \{x\}$ is infinite, hence closed. Therefore, for any subset $S \subseteq X$, $$ S = \bigcup_{x \in S} \{x\} $$ is a union of open sets, thus open. This means $T$ is all subsets of $X$. You have the discrete topology.
Hope that helps,