Topological space inducing a space such that it is a subspace of that space? If $\;Y\subseteq X^2$ and $\tau$ is any topology on $Y$, is it possible to induce topologies in $X$ from $\tau$?
Are there ways to define topologies on $X$ from $\tau$ so that $(Y,\tau)$ become a subspace of the product space?

Erroneous
Define $\tau_1,\;\tau_2$ on $X$ by the two functions $p_k:Y\to X$ defined as $p_k(x_1,x_2)=x_k$:


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*Given $A\subseteq X,\;x\in\bar A\iff   
    p_k^{-1}(x)\subseteq\overline{p_k^{-1}(A)}, \;k=1,2$.
(NOT TRU! WILL REPAIR OR DELETE)


I used this unauthorized method to prove 1: any relation 
$\propto\,\subseteq X\times \mathcal P(X)$ that extends $'\!\!\in'$ in the way that, 


*

*$x\in M\Rightarrow x\propto M$

*$\neg\exists x\in X:x\propto\emptyset$

*$x\propto A \subseteq B \Rightarrow x\propto B$

*$x\propto A\cup B\Rightarrow x\propto A \vee x\propto B$


defines a closure operation on subsets of $X$ by 
$x\in \overline M \Leftrightarrow x\propto M$,
 A: Let $X=\{0,1\}$, and let $Y=X\times X$. Let 
$$\mathscr{B}=\big\{\{\langle 0,0\rangle\},\{\langle 1,0\rangle\},X\times\{1\}\big\}\;,$$
and let $\tau$ be the topology on $Y$ generated by the base $\mathscr{B}$. 
Suppose that there topologies $\tau_1$ and $\tau_2$ on $X$ such that $Y$ is the product topology on $X\times X$ generated by $\tau_1$ and $\tau_2$; the projection map $\pi_1$ to the first coordinate must be be continuous and open. Since $\pi_1^{-1}[\{0\}]$ and $\pi_1^{-1}[\{1\}]$ are not open in $Y$, continuity of $\pi_1$ implies that $\{0\},\{1\}\notin\tau_1$ and hence that $\tau_1=\{\varnothing,X\}$, the indiscrete topology on $X$. On the other hand, the fact that $\pi_1$ is open implies that $\{0\}=\pi_1[\{\langle 0,0\rangle\}]$ and $\{1\}=\pi_1[\{\langle 1,0\rangle\}]$ are open in $\langle X,\tau_1\rangle$ and hence that $\tau_1$ is the discrete topology $\wp(X)$ on $X$. This contradiction shows that no such topologies $\tau_1$ and $\tau_2$ exist: $Y$ is not (a subspace of) any product topology on $X\times X$.
A: In general, if $X$ is a set, $Y$ a topological space and $f : Y \rightarrow X$ is a function, then we can define $\tau_{f} = \{U \subseteq X \:|\: f^{-1}(U) \in \tau_Y\}$ where $\tau_Y$ is the topology of $Y$. Then $\tau_f$ is a topology of $X$, in fact, it is the finest topology for which $f$ is continuous. 
In your situation, if $Y \subseteq X^2$ is endowed with a topology $\tau_Y$, let $p_k : Y \rightarrow X$ be the restrictions of the two projections of $X^2$ to $X$. Then $\tau_{p_k}$ gives a topology on $X$ for which $p_k$ is continuous. Hence the inclusion map $p_1 \times p_2 : Y \rightarrow X^2$ is continuous when $X^2$ is equipped with the product topology $\tau_{p_1} \times \tau_{p_2}$ and so $Y$ is a subspace. Note also that this product topology is the same as the topology $\tau_{p_1 \times p_2}$, so a subset of $X^2$  is open in this topology if and only if the intersection of this set with $Y$ is open in $Y$. This topology is the finest topology such that $Y$ is a subspace of $X^2$.
