The property you are asking for is a very strong condition on $T$. Let met try to rephrase the question more carefully:

The set of countable $\Sigma$-structures with universe $\omega$ is naturally a Polish space $\mbox{Mod}(\Sigma)$ which has no isolated points (if it has isolated points then necessarily $\Sigma$ would consist of a finite set of constant symbols, contradicting the existence of $T$ with continuum many countable models). I assume that the ``reasonable way of coding reals as $\Sigma$-structures" is some Borel bijection $f: \mathbb{R} \to \mbox{Mod}(\Sigma)$; for your question, choosing such a Borel bijection is the same thing as fixing a new completely metrizable topology on $\mbox{Mod}(\Sigma)$ (which induces the same Borel structure on $\mbox{Mod}(\Sigma)$ as the original topology).

Now $\mbox{Mod}(T)$ is a Borel subset of $\mbox{Mod}(\Sigma)$; we are supposing that $\mbox{Mod}(T)/\cong$ has size continuum. It doesn't automatically follow that there is a perfect subset $P$ of $\mbox{Mod}(T)$ of pairwise nonisomorphic models (if $2^{\aleph_0} = \aleph_1$ then this is Vaught's conjecture).

In any case, we are asking for a perfect set $P$ of $\mbox{Mod}(T)$ of pairwise nonisomorphic models, such that every element of $\mbox{Mod}(T)$ is isomorphic to exactly one element of $P$. In the terminology of invariant descriptive set theory, this is asking for a perfect transversal of $(\mbox{Mod}(T), \cong)$. In general, even the existence of a Borel transversal of $(\mbox{Mod}(T), \cong)$ is quite strong:

Note that first of all, if $(\mbox{Mod}(T), \cong)$ has a Borel transversal $S$ than $\cong$ is Borel, since $M \not \cong N$ iff $\exists M', N' \in S$ with $M \cong M', N \cong N'$ and $M' \not= N'$. Hence by Theorem 6.4.4 from Su Gao's textbook on Invariant Descriptive Set Theory, if $(\mbox{Mod}(T), \cong)$ has a Borel transversal then $(\mbox{Mod}(T), \cong)$ is smooth, that is $(\mbox{Mod}(T), \cong) \leq_B (\mathbb{R}, =)$.

In terms of Borel complexity theory this is saying that if $(\mbox{Mod}(T), \cong)$ has a Borel transversal then it is at the bottom of the complexity hierarchy. For instance, any non-small theory is not smooth.

The property you are asking for is a very strong condition on $T$. Let met try to rephrase the question more carefully:

The set of countable $\Sigma$-structures with universe $\omega$ is naturally a Polish space $\mbox{Mod}(\Sigma)$ which has no isolated points (if it has isolated points, then necessarily $\Sigma$ would consist of a finite set of constant symbols, contradicting the existence of $T$ with continuum many countable models). I assume that the ``reasonable way of coding reals as $\Sigma$-structures" is some Borel bijection $f: \mathbb{R} \to \mbox{Mod}(\Sigma)$; for your question, choosing such a Borel bijection is the same thing as fixing a new completely metrizable topology on $\mbox{Mod}(\Sigma)$ (which induces the same Borel structure on $\mbox{Mod}(\Sigma)$ as the original topology).

Now $\mbox{Mod}(T)$ is a Borel subset of $\mbox{Mod}(\Sigma)$; we are supposing that $\mbox{Mod}(T)/\cong$ has size continuum. It doesn't automatically follow, that there is a perfect subset $P$ of $\mbox{Mod}(T)$ of pairwise nonisomorphic models (if $2^{\aleph_0} = \aleph_1$, then this is Vaught's conjecture).

In any case, we are asking for a perfect set $P$ of $\mbox{Mod}(T)$ of pairwise nonisomorphic models, such that every element of $\mbox{Mod}(T)$ is isomorphic to exactly one element of $P$. In the terminology of invariant descriptive set theory, this is asking for a perfect transversal of $(\mbox{Mod}(T), \cong)$. In general, even the existence of a Borel transversal of $(\mbox{Mod}(T), \cong)$ is quite strong:

Note that first of all, if $(\mbox{Mod}(T), \cong)$ has a Borel transversal $S$ then $\cong$ is Borel, since $M \not \cong N$ iff $\exists M', N' \in S$ with $M \cong M', N \cong N'$ and $M' \not= N'$. Hence by Theorem 6.4.4 from Su Gao's textbook on Invariant Descriptive Set Theory, if $(\mbox{Mod}(T), \cong)$ has a Borel transversal, then $(\mbox{Mod}(T), \cong)$ is smooth, that is $(\mbox{Mod}(T), \cong) \leq_B (\mathbb{R}, =)$.

In terms of Borel complexity theory this is saying, that if $(\mbox{Mod}(T), \cong)$ has a Borel transversal, then it is at the bottom of the complexity hierarchy. For instance, any non-small theory is not smooth.

The property you are asking for is a very strong condition on $T$. Let met try to rephrase the question more carefully:

The set of countable $\Sigma$-structures with universe $\omega$ is naturally a Polish space $\mbox{Mod}(\Sigma)$ which has no isolated points (if it has isolated points then necessarily $\Sigma$ would consist of a finite set of constant symbols, contradicting the existence of $T$ with continuum many countable models). I assume that the ``reasonable way of coding reals as $\Sigma$-structures" is some Borel bijection $f: \mathbb{R} \to \mbox{Mod}(\Sigma)$; for your question, choosing such a Borel bijection is the same thing as fixing a completely metrizable topology on $\mbox{Mod}(\Sigma)$.

Now $\mbox{Mod}(T)$ is a Borel subset of $\mbox{Mod}(\Sigma)$; we are supposing that $\mbox{Mod}(T)/\cong$ has size continuum. It doesn't automatically follow that there is a perfect subset $P$ of $\mbox{Mod}(T)$ of pairwise nonisomorphic models (if $2^{\aleph_0} = \aleph_1$ then this is Vaught's conjecture).

In any case, we are asking for a perfect set $P$ of $\mbox{Mod}(T)$ of pairwise nonisomorphic models, such that every element of $\mbox{Mod}(T)$ is isomorphic to exactly one element of $P$. In the terminology of invariant descriptive set theory, this is asking for a perfect transversal of $(\mbox{Mod}(T), \cong)$. In general, even the existence of a Borel transversal of $(\mbox{Mod}(T), \cong)$ is quite strong:

Note that first of all, if $(\mbox{Mod}(T), \cong)$ has a Borel transversal $S$ than $\cong$ is Borel, since $M \not \cong N$ iff $\exists M', N' \in S$ with $M \cong M', N \cong N'$ and $M' \not= N'$. Hence by Theorem 6.4.4 from Su Gao's textbook on Invariant Descriptive Set Theory, if $(\mbox{Mod}(T), \cong)$ has a Borel transversal then $(\mbox{Mod}(T), \cong)$ is smooth, that is $(\mbox{Mod}(T), \cong) \leq_B (\mathbb{R}, =)$.

In terms of Borel complexity theory this is saying that if $(\mbox{Mod}(T), \cong)$ has a Borel transversal then it is at the bottom of the complexity hierarchy. For instance, any non-small theory is not smooth.

The property you are asking for is a very strong condition on $T$. Let met try to rephrase the question more carefully:

The set of countable $\Sigma$-structures with universe $\omega$ is naturally a Polish space $\mbox{Mod}(\Sigma)$ which has no isolated points (if it has isolated points then necessarily $\Sigma$ would consist of a finite set of constant symbols, contradicting the existence of $T$ with continuum many countable models). I assume that the ``reasonable way of coding reals as $\Sigma$-structures" is some Borel bijection $f: \mathbb{R} \to \mbox{Mod}(\Sigma)$; for your question, choosing such a Borel bijection is the same thing as fixing a new completely metrizable topology on $\mbox{Mod}(\Sigma)$ (which induces the same Borel structure on $\mbox{Mod}(\Sigma)$ as the original topology).

Now $\mbox{Mod}(T)$ is a Borel subset of $\mbox{Mod}(\Sigma)$; we are supposing that $\mbox{Mod}(T)/\cong$ has size continuum. It doesn't automatically follow that there is a perfect subset $P$ of $\mbox{Mod}(T)$ of pairwise nonisomorphic models (if $2^{\aleph_0} = \aleph_1$ then this is Vaught's conjecture).

In any case, we are asking for a perfect set $P$ of $\mbox{Mod}(T)$ of pairwise nonisomorphic models, such that every element of $\mbox{Mod}(T)$ is isomorphic to exactly one element of $P$. In the terminology of invariant descriptive set theory, this is asking for a perfect transversal of $(\mbox{Mod}(T), \cong)$. In general, even the existence of a Borel transversal of $(\mbox{Mod}(T), \cong)$ is quite strong:

Note that first of all, if $(\mbox{Mod}(T), \cong)$ has a Borel transversal $S$ than $\cong$ is Borel, since $M \not \cong N$ iff $\exists M', N' \in S$ with $M \cong M', N \cong N'$ and $M' \not= N'$. Hence by Theorem 6.4.4 from Su Gao's textbook on Invariant Descriptive Set Theory, if $(\mbox{Mod}(T), \cong)$ has a Borel transversal then $(\mbox{Mod}(T), \cong)$ is smooth, that is $(\mbox{Mod}(T), \cong) \leq_B (\mathbb{R}, =)$.

In terms of Borel complexity theory this is saying that if $(\mbox{Mod}(T), \cong)$ has a Borel transversal then it is at the bottom of the complexity hierarchy. For instance, any non-small theory is not smooth.