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The possibly unsatisfying answer to your question is "sometimes." I will instead discuss the obviously equivalent question about universal closed subsets (it will let me use more standard notation later). Moreover, I will focus on the special case that $X$ and $Y$ are both Polish, since that has been examined more in the literature.

First, let me point out an oversight in your analysis of the case that $X$ is finite. Certainly $X$ must have the discrete topology, so every subset of $X$ is closed. However, $Y = \mathcal{P}(X)$ is a perfectly fine Polish space when endowed with its own discrete topology. Then the set $\{(x,A) \in X \times \mathcal{P}(X) : x \in A\}$ is "uniquely" universal closed.

This may seem pedantic, but it actually generalizes to large $X$. Suppose now that $X$ is a compact Polish space, and endow its space of compact (equiv., closed) subsets $\mathcal{K}(X)$ with the Vietoris topology, generated by sets of the form

$\{K : K \subseteq U\}$ and $\{K : K \cap U = \emptyset\}$,

where $U\subseteq X$ is open. For Polish $X$, this is a Polish topology on $\mathcal{K}(X)$. Note that in the special case where $X$ is finite (thus compact), this coincides with the discrete topology on $\mathcal{P}(X)$. Motivated by this analogy, we proceed as before and choose our uniquely universal closed set to equal $G = \{(x,K) \in X \times \mathcal{K}(X) : x \in K\}$. The only thing left to check is that this set is indeed closed. You can see this directly by assuming $(x_0,K_0) \notin G$, fixing a little open neighborhood $U$ around $x_0$ disjoint from $K_0$, and then checking that $U \times \{K : K \cap U = \emptyset\}$ is an open neighborhood of $(x_0, K_0)$ disjoint from $G$.

The obvious place to look for more information about this is Kechris' descriptive set theory text. Unfortunately I don't have a copy on hand at the moment (which makes me feel like a child without a security blanket), so I can't give more specific references.

Moving on. For noncompact Polish spaces $X$ you can endow the space $\mathrm{CL}(X)$ of closed subsets of $X$ with a topology called the Wijsman topology. Well, really there are several such topologies, since the definition relies on a choice of compatible complete metric $d$ on $X$. This topology is the weakest topology making the functions $f_x : A \mapsto d(x, A)$ continuous for each $x \in X$. It is a result of Gerald Beer's that this topology is Polish for $(X,d)$ as above. (This might well be in Kechris' book, but as I mentioned I don't have it on hand so I'll regurgitate the reference that google gave me.)

Beer, Gerald. A Polish topology for the closed subsets of a Polish space. Proc. Amer. Math. Soc. 113 (1991), no. 4, 1123–1133.

Edit: but Theo Buehler can give has given a relevant reference to Kechris! . See his comment.

A variation of the earlier argument in the compact case should work in this context.

Edit again: I just noticed that the definition of the topology I gave makes sense for nonempty closed subsets of $X$. This is not a serious problem and is in fact addressed in Beer's paper.

Finally, it is hopeless to expect this to work for arbitrary Polish spaces $X$ and $Y$. As you noticed, for small spaces there are cardinality issues. When the spaces are large, you can also fiddle around with compactness/noncompactness, and other topological notions. There are just too many wild Polish spaces.
show/hide this revision's text 2 grammar, added reference, changed unnecessary sentence on projective hierarchy; added 37 characters in body

The possibly unsatisfying answer to your question is "sometimes." I will instead discuss the obviously equivalent question about universal closed subsets (it will let me use more standard notation later). Moreover, I will focus on the special case that $X$ and $Y$ are both Polish, since that has been examined on more in the literature.

First, let me point out an oversight in your analysis of the case that $X$ is finite. Certainly $X$ must have the discrete topology, so every subset of $X$ is closed. However, $Y = \mathcal{P}(X)$ is a perfectly fine Polish space when endowed with its own discrete topology. Then the set $\{(x,A) \in X \times \mathcal{P}(X) : x \in A\}$ is "uniquely" universal closed.

This may seem pedantic, but it actually generalizes to large $X$. Suppose now that $X$ is a compact Polish space, and endow its space of compact (equiv., closed) subsets $\mathcal{K}(X)$ with the Vietoris topology, generated by sets of the form

$\{K : K \subseteq U\}$ and $\{K : K \cap U = \emptyset\}$,

where $U\subseteq X$ is open. For Polish $X$, this is a Polish topology on $\mathcal{K}(X)$. Note that in the special case where $X$ is finite (thus compact), this coincides with the discrete topology on $\mathcal{P}(X)$. Motivated by this analogy, we proceed as before and choose our uniquely universal closed set to equal $G = \{(x,K) \in X \times \mathcal{K}(X) : x \in K\}$. The only thing left to check is that this set is indeed closed. You can see this directly by assuming $(x_0,K_0) \notin G$, fixing a little open neighborhood $U$ around $x_0$ disjoint from $K_0$, and then checking that $U \times \{K : K \cap U = \emptyset\}$ is an open neighborhood of $(x_0, K_0)$ disjoint from $G$.

The obvious place to look for more information about this is Kechris' descriptive set theory text. Unfortunately I don't have a copy on hand at the moment (which makes me feel like a child without a security blanket), so I can't give more specific references.

Moving on. For noncompact Polish spaces $X$ you can endow the space $\mathrm{CL}(X)$ of closed subsets of $X$ with a topology called the Wijsman topology. Well, really there are several such topologies, since the definition relies on a choice of compatible complete metric $d$ on $X$. This topology is the weakest topology making the functions $f_x : A \mapsto d(x, A)$ continuous for each $x \in X$. It is a result of Gerald Beer's that this topology is Polish for $(X,d)$ as above. (This might well be in Kechris' book, but as I mentioned I don't have it on hand so I'll regurgitate the reference that google gave me.)

Beer, Gerald. A Polish topology for the closed subsets of a Polish space. Proc. Amer. Math. Soc. 113 (1991), no. 4, 1123–1133.

Edit: but Theo Buehler can give a reference to Kechris! See his comment.

A variation of the earlier argument in the compact case should work in this context.

Finally, it is hopeless to expect this to work for arbitrary Polish spaces $X$ and $Y$. As you noticed, for small spaces there are cardinality issues. When the spaces are large, you can also fiddle around with compactness/noncompactness, and other topological notions. There are just too many wild Polish spaces. Fortunately, once you climb up the projective hierarchy a bit, these topological issues vanish since there's only one standard Borel space.
show/hide this revision's text 1

The possibly unsatisfying answer to your question is "sometimes." I will instead discuss the obviously equivalent question about universal closed subsets (it will let me use more standard notation later). Moreover, I will focus on the special case that $X$ and $Y$ are both Polish, since that has been examined on more in the literature.

First, let me point out an oversight in your analysis of the case that $X$ is finite. Certainly $X$ must have the discrete topology, so every subset of $X$ is closed. However, $Y = \mathcal{P}(X)$ is a perfectly fine Polish space when endowed with its own discrete topology. Then the set $\{(x,A) \in X \times \mathcal{P}(X) : x \in A\}$ is "uniquely" universal closed.

This may seem pedantic, but it actually generalizes to large $X$. Suppose now that $X$ is a compact Polish space, and endow its space of compact (equiv., closed) subsets $\mathcal{K}(X)$ with the Vietoris topology, generated by sets of the form

$\{K : K \subseteq U\}$ and $\{K : K \cap U = \emptyset\}$,

where $U\subseteq X$ is open. For Polish $X$, this is a Polish topology on $\mathcal{K}(X)$. Note that in the special case where $X$ is finite (thus compact), this coincides with the discrete topology on $\mathcal{P}(X)$. Motivated by this analogy, we proceed as before and choose our uniquely universal closed set to equal $G = \{(x,K) \in X \times \mathcal{K}(X) : x \in K\}$. The only thing left to check is that this set is indeed closed. You can see this directly by assuming $(x_0,K_0) \notin G$, fixing a little open neighborhood $U$ around $x_0$ disjoint from $K_0$, and then checking that $U \times \{K : K \cap U = \emptyset\}$ is an open neighborhood of $(x_0, K_0)$ disjoint from $G$.

The obvious place to look for more information about this is Kechris' descriptive set theory text. Unfortunately I don't have a copy on hand at the moment (which makes me feel like a child without a security blanket), so I can't give more specific references.

Moving on. For noncompact Polish spaces $X$ you can endow the space $\mathrm{CL}(X)$ of closed subsets of $X$ with a topology called the Wijsman topology. Well, really there are several such topologies, since the definition relies on a choice of compatible complete metric $d$ on $X$. This topology is the weakest topology making the functions $f_x : A \mapsto d(x, A)$ continuous for each $x \in X$. It is a result of Gerald Beer's that this topology is Polish for $(X,d)$ as above. (This might well be in Kechris' book, but as I mentioned I don't have it on hand so I'll regurgitate the reference that google gave me.)

Beer, Gerald. A Polish topology for the closed subsets of a Polish space. Proc. Amer. Math. Soc. 113 (1991), no. 4, 1123–1133.

A variation of the earlier argument in the compact case should work in this context.

Finally, it is hopeless to expect this to work for arbitrary Polish spaces $X$ and $Y$. As you noticed, for small spaces there are cardinality issues. When the spaces are large, you can also fiddle around with compactness/noncompactness, and other topological notions. There are just too many wild Polish spaces. Fortunately, once you climb up the projective hierarchy a bit, these topological issues vanish since there's only one standard Borel space.