How do I prove that the product topology is the unique topology for which projections are well-behaved?

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This is a question from a practice qualifying exam.

Let $X, Y, T$ be topological spaces. Define $p_X: X\times Y\to X$ and $p_Y:X\times Y\to Y$ to be the projection maps. Then $f:T\to X\times Y$ is continuous if and only if the compositions $p_X \circ f : T\to X$ and $p_Y\circ f: T\to Y$ are continuous. (I have already proven this).

Prove that the product topology on $X\times Y$ is the unique topology that, for all spaces $T$ and functions $f$, has the previous property.

As shown here: Product topology on $X \times Y$ the smallest topology when $f(x, y) = x$ and $g(x, y) = y$ are continuous functions?, we know that any topology n $X\times Y$ with the property must at least contain the product topology (just use $T = X\times Y$ and $f = id$). Thus, the only thing that I need to show is that any topology for which the property holds is contained in the product topology, but unfortunately I'm not really sure how to approach this.

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First, note that if you have any topology $\tau$ on $X\times Y$ with this property, then the projections $p_X$ and $p_Y$ are both continuous with respect to $\tau$ (take $T=(X\times Y,\tau)$ and $f=id$). Now if $\tau_1$ and $\tau_2$ are two topologies on $X\times Y$ with this property, then the identity map $X\times Y\to X\times Y$ is continuous from $\tau_1$ to $\tau_2$ (use the property for $\tau_2$ and take $T=(X\times Y,\tau_1)$, $f=id$). By symmetry, the identity map is also continuous from $\tau_1$ to $\tau_2$. Thus $\tau_1=\tau_2$.