$\Theta \rho ϵ\eta \Pi \alpha \tau \pi$

$⟸$ If it interestects in infinitely many points then there is a point not equal to $x$ where it intersects so that $x\in A^{\prime }$

$⟹$ Suppose that $x\in A^{\prime }$ so that $x\in \overline{A\setminus \left\{x\right\}}$ now take $U$; a neighborhood of $x$ therefore $U\cap (A\setminus \left\{x\right\})\ne \varnothing$ so take a point $a_1$ in there and consider $U_1=U\setminus \left\{a_1\right\}$, since $X$ is $T_1$ then we know that $U_1$ is open and still a neighborhood of $x$ and so by induction we will construct an $a_2,a_3,\dots$ all of which are elements of $U\cap A$ so that it contains infinitely many points of $A$

Suppose that $x_n\to a$ and $x_n\to b$ for $a\ne b\in X$ then we obtain neighborhoods $U_a,U_b$ such that $a\in U_a$ and $b\in U_B$ and that $U_a\cap U_b=\varnothing$, but now $U_a$ contains all but finitely many elements in the sequence, therefore there are only finitely many elements of the sequence in $X\setminus U_a\supseteq U_b$ which is a contradiction because $U_b$ also contains all bute finitely many elements of the sequence.

We will show that $\left\{p\right\}$ is closed by showing $\left\{p\right\}=\overline{\left\{p\right\}}$

Suppose that $y\in X$ and $y\ne p$, since $X$ was Hausdorff, we get $U_p,U_y$ disjoint neighborhoods. Since $U_y\cap U_p=\varnothing$ , and $U_p\supseteq \left\{p\right\}$, then $U_y\cap \left\{p\right\}=\varnothing$ therefore $y\notin \overline{\left\{p\right\}}$. Note that we've just shown that for any point in $X$ which is not $p$ it cannot be in the closure.

Now we show that $p$ is actually in there. Since $\left\{p\right\}\subseteq \overline{\left\{p\right\}}$, then we know $p\in \overline{\left\{p\right\}}$, so therefore $\overline{\left\{p\right\}}=\left\{p\right\}$ as needed.

Suppose that $\mathrm{\Delta }$ is closed in $X\times X$, now let $x\ne y\in X$, if we consider the point $p=(x,y)\in X\times X$ then we know that $p\notin \mathrm{\Delta }$, therefore $p\in X\times X\setminus \mathrm{\Delta }$, since $\mathrm{\Delta }$ was closed, then we know that $X\times X\setminus \mathrm{\Delta }$ is open, and therefore since $p$ is in there, then there exists a basis element $U\times V$ where $U,V$ are open in $X$ which contains $p$ and is a subset of $X\times X\setminus \mathrm{\Delta }$. To show that the space is hausdorff it remains to show that $U\cap V=\varnothing$, so suppose for the sake of contradiction that there was an element $d\in U\cap V$ which implies that since $(d,d)\in U\times V$ but $U\times V\subseteq X\times X\setminus \mathrm{\Delta }$ which shows that such an element $d$ is impossible to exist, and therefore we must have $U\cap V=\varnothing$

Now suppose that $X$ is hausdorff, let's prove that $\mathrm{\Delta }$ is closed in $X\times X$, so we will prove that $X\times X\setminus \mathrm{\Delta }$ is open, so let $p\in X\times X\setminus \mathrm{\Delta }$, we have $p=(x,y)$ where $x\ne y\in X$ therefore since $X$ is hausdorff then we get open neighborhoods $U,V$ of $x,y$ respectively that are disjoint, so that $U\cap V=\varnothing$ this shows that $(U\times V)\cap \mathrm{\Delta }=\varnothing$ as if it was not empty, then $U,V$ would no longer be disjoint, therefore we know that $p\in U\times V\subseteq X\times X\setminus \mathrm{\Delta }$, so that $X\times X\setminus \mathrm{\Delta }$ is open and thus $\mathrm{\Delta }$ is closed, as needed.

For the base case of $n=2$ it follows directly from the definition of a hausdorff space, now let $k\in {\mathbb{N}}_2$ and suppose that the statement holds true on $k$ elements, now we'll show that it holds true on $k+1$ elements. Let $C=x_1,\dots ,x_{k+1}$ be the $k+1$ elements, let $C_i=C\setminus \left\{x_i\right\}$, which produces a set of size $k$ on which we know the induction hypothesis holds true, and therefore we obtain $U_{1,i},\dots ,U_{k,i}$ disjoint open sets such that $x_j\in U_{j,i}$ for all $j\ne i$ and lets call the family of these sets by the mask of $C_i$

By iterating $i$ over $1,\dots k+1$ then we can form smaller open sets around each $x_j$ by taking ${\bigcap }_i{U}_{j,i}$, they are open as they are finite intersections of open sets, and we can show that they pairwise disjoint, for consider any $a\ne b\in [1,\dots ,n]$ and consider ${\bigcap }_i{U}_{a,i}$ and ${\bigcap }_i{U}_{b,i}$, since $k+1\ge 3$ then we can consider some $m\in [1,\dots ,k+1]$ such that $m\ne a$ and $m\ne b$ then consider the $m$-th mask, $U_{1,m},\dots U_{k,m}$ of $C_m$ as noted in the previous paragraph the induction hypothesis shows these are pairwise disjoint, additionally we know that ${\bigcap }_i{U}_{a,i}\subseteq U_{a,m}$ and that ${\bigcup }_i{U}_{b,i}\subseteq U_{b,m}$ (this is because they are part of the intersection), now since $U_{a,m}$ and $U_{b,m}$ are disjoint then are the intersections in question, thus we've shown that the sets $\{x\in [1,\dots ,k+1]:{\bigcup }_i{U}_{x,i}\}$ are pairwise disjoint and cover $x_1,\dots ,x_{k+1}$, as needed, therefore by the principle of induction this holds true for all $n\in {\mathbb{N}}_2$

Suppose $a=(x_a,y_a)$ and $b=(x_b,y_b)$ and assume that $a\ne b$ so without loss of generality assume that $x_a\ne x_b$ therefore we obtain disjoint neighborhoods of each $U_a,U_b$ then $U_a\times Y$ and $U_b\times Y$ are disjoint neighborhoods of $a,b$ respectively.