my attempt for -----option A : Let X be Hausdorff, then if $x\ne y$ there are neighborhoods $V_x$ and $V_y$ such that $V_x \cap V_y = \emptyset$. Therefore $V_x\times V_y \cap D=\emptyset$ and the complement of $D$ is open. Now, assume that the latter is true. Then, for any point $(x,y)$, $x\ne y$, there is an open set around it that does not intersect $D$. Therefore, there are two sets $x\in V_x$ and $y\in V_y$ such that $V_x\times V_y$ doesn't intersect $D$, therefore $V_x \cap V_y = \emptyset$ ....as X is Hausdorff since the diagonal is closed.
I don't know how to prove OPtion B and option C pliz help me
I would be more grate ful
thanks in advances
A)
Let $x,y\in X$ with $x\neq y$ or equivalently $\langle x,y\rangle\in d^{-1}(\{0\})^{\complement}=d^{-1}((0,\infty))\subseteq X\times X$.
For proving that $(X,\tau)$ is Hausdorff it is enough to show that sets $U,V\in\tau$ exist with $x\in U$, $y\in V$ and $U\cap V=\varnothing$.
Here $(0,\infty)$ is open and $d$ is continuous so that $d^{-1}((0,\infty))$ is open.
Further $X\times X$ is equipped with the product topology so $U,V\in\tau$ exist with $$\langle x,y\rangle\subseteq U\times V\subseteq d^{-1}((0,\infty))$$
Then $U$ and $V$ satisfy the mentioned conditions and we are ready.
B)
To be shown is here that for every pair $B_1,B_2$ of such sets and every $y\in B_1\cap B_2$ there is a another such set $B$ with $y\in B\subseteq B_1\cap B_2$, and further that for every $y$ such a set $B$ exists with $y\in B$.
It is unclear to me whether we are dealing with a basis $\{B_{x,\epsilon}\mid\epsilon>0\}$ for some fixed $x$ or with a basis $\{B_{x,\epsilon}\mid\epsilon>0, x\in X\}$ and I will leave this to you.
C)
For fixed $x\in X$ prescribe function $f_x:X\to X\times X$ by $y\mapsto\langle x,y\rangle$. If $p_1$ and $p_2$ denote the projections then $p_1\circ f:X\to X$ is constant and $p_2\circ f=\mathsf{id}_X$. So both functions are continuous and the fact that $X\times X$ is equipped with product topology allows the conclusion that $f_x:X\to X\times X$ is continuous. Then composition $d\circ f:X\to X$ is continuous so that $(d\circ f)^{-1}([0,\epsilon))\in\tau$ for $\epsilon>0$. Now observe that: $$B(x,\epsilon)=(d\circ f)^{-1}([0,\epsilon))$$So actually it has been shown that $\tau$ contains a basis of $\tau'$. Then also $\tau'\subseteq\tau$ as was to be shown.