Cantor Set (Hausdorff)

Being a subspace of the metric space $\mathbb R$, the Cantor set is metrizable. This implies that the Cantor set satisfies all the usual separation axioms --- Hausdorff, regular, completely regular, normal, hereditarily normal, perfectly normal.

"I know that R (real numbers) are Hausdor[f]f is it enough?"

Yes, it is. Suppose $X$ is a Hausdorff space and $Y\subseteq X$ has the subspace topology.

Given two points $x,y\in Y,$ there are open sets $U,V\subseteq X$ with $x\in U,$ $y\in V,$ and $U\cap V=\varnothing.$ The intersections of $U$ and $V$ with $Y$ are open subsets of $Y$ and contain $x$ and $y$ as members and do not intersect each other.

Given any two points $p,q\in \mathfrak{C}$ (the Cantor Ternary Set), we have that $p,q\in \mathbf{R}$ as well. Take open neighborhoods $U,V\subset \mathbf{R}$ such that $p\in U$, $q\in V$ and $U\cap V=\varnothing$. Then if we take intersections $\tilde{U}=\mathfrak{C}\cap U$ and $\tilde{V}=\mathfrak{C}\cap V$, we have that $p\in \tilde{U}$ and $q\in \tilde{V}$, with $\tilde{U}\cap \tilde{V}=\varnothing$. By definition of the subspace topology, $\tilde{U}$ and $\tilde{V}$ are open in $\mathfrak{C}$ and we are done.

Note that this construction works in general, for any subspace $Y$ of a Hausdorff space $X$.