A space whose powers "generate" all spaces via quotients I've been asked to prove (or disprove) that there exists a topological space $X$ with the following property:

For every space $T$ there is a set $I$ and a homeomorphism $T\cong A$ where $A$ is a quotient of a subspace of $X^I = \prod_{i\in I} X$ (product topology)

Of course, this smells like a universal property, and a rather strong one, so I'm inclined to think there is no such $X$. But how can one prove that such a space does not exist?

You can do this without even needing quotients.  Let $X=\{0,0',1\}$, with the topology $\{X,\emptyset,\{1\}\}$.  Given any space $T$, let $I$ be the set of all continuous maps $T\to X$.  There is a canonical continuous map $F:T\to X^I$ given by $F(t)(f)=f(t)$.  I claim that $F$ is a homeomorphism onto its image, so $T$ is homeomorphic to the subspace $A=F(T)\subseteq X^I$.
First, $F$ is injective, since the subspace $\{0,0'\}$ of $X$ is indiscrete and so any map $T\to \{0,0'\}$ is continuous.  In particular, given any two distinct points of $T$, we can find a continuous map $T\to X$ which sends one of them to $0$ and the other to $0'$.
Second, $F$ is open as a map onto its image.  Indeed, let $U\subseteq T$ be open.  Then the function $f:T\to X$ sending $U$ to $1$ and $T\setminus U$ to $0$ is continuous.  The image $F(U)$ is then exactly the set of elements of $F(T)$ whose $f$ coordinate is $1$.  This is an open subset of $F(T)$ since $\{1\}$ is open in $X$.  Thus $F$ is open as a map to $F(T)$.
