Sets,transversals,PT property,cardinals A transversal of a family $S$ of sets is an injective choice function.
$PT(\lambda,\chi)$ means, if $S$ is a family of $\lambda$ sets,each of cardinality $<\chi$,and every subfamily with $<\lambda$ sets has a transversal, then $S$ has a transversal.
My questions tackles only the border of the problem: does $$PT(\aleph_0,\aleph_0)$$ hold?
Why not $PT(\aleph_1,\aleph_1)$?
 A: To answer your second question, recall that sets $A,B\subseteq\Bbb N$ are almost disjoint if $A\cap B$ is finite. It’s well known that there is an almost disjoint family $\mathscr{D}\subseteq\wp(\Bbb N)$ such that $|\mathscr{D}|=2^\omega=\mathfrak{c}$. Let $\mathscr{D}_0$ be any subset of $\mathscr{D}$ of cardinality $\omega_1$. Clearly $\mathscr{D}_0$ has no transversal, since $\bigcup\mathscr{D}_0$ is countable. However, if $\mathscr{A}=\{D_k:k\in\omega\}\subseteq\mathscr{D}_0$, it’s easy recursively to construct a transversal $f$ for $\mathscr{A}$: given $f(k)\in D_k$ for $k<m$, let $f(m)=\min\left(D_m\setminus\bigcup_{k<m}D_k\right)$. (Here we use the fact that $\mathscr{D}_0$ is an almost disjoint family: $\bigcup_{k<m}(D_m\cap D_k)$ is finite, so $D_m\setminus\bigcup_{k<m}D_k$ is infinite and in particular non-empty.)
For the first question let $\mathscr{S}=\{S_\alpha:\alpha\in A\}$ be a family of finite sets such that every finite subfamily of $\mathscr{S}$ has a transversal. (I don’t even need to require the index set $A$ to be countable.) Endow each $S_\alpha$ with the discrete topology, and let $X=\prod_{\alpha\in A}S_\alpha$ with the product topology. 
For finite $F\subseteq A$ let $\mathscr{S}_F=\{S_\alpha:\alpha\in F\}$ and
$$\begin{align*}
H_F&=\{x\in X:x(\alpha)\ne x(\beta)\text{ whenever }\alpha,\beta\in F\text{ and }\alpha\ne\beta\}\\
&=\{x\in X:x\upharpoonright F\text{ is a transversal for }\mathscr{S}_F\}\;.
\end{align*}$$
$H_F$ is a closed set in $X$, and by hypothesis $H_F\ne\varnothing$. Let $\mathscr{H}=\{H_F:F\subseteq A\text{ is finite}\}$. If $F,G\subseteq A$ are finite, then $H_F\cap H_G=H_{F\cup G}\in\mathscr{H}$, so $\mathscr{H}$ has the finite intersection property. $X$ is compact by the Tikhonov product theorem, so $\bigcap\mathscr{H}\ne\varnothing$, and any $x\in\bigcap\mathscr{H}$ is a transversal for $\mathscr{S}$.
Added: The preceding argument can be recast as an application of Zorn’s lemma. Let $S=\bigcup\mathscr{S}$. For $C\subseteq A$ say that a function $f:C\to S$ is good if for every finite $F\subseteq A$ the restriction $f\upharpoonright(C\cap F)$ is a transversal for $\mathscr{S}_{C\cap F}$, and let
$$\mathscr{P}=\{\langle D,s\rangle:D\subseteq A\text{ and }s:D\to S\text{ is good}\}\;.$$
You should check that whenever $D\subseteq A$ and $s:D\to S$ is good, then $s$ is a transversal for $\mathscr{S}_D$; this is very straightforward.
Define a partial order $\preceq$ on $\mathscr{P}$ as follows: for $\langle D,s\rangle,\langle E,t\rangle\in\mathscr{P}$ set $\langle D,s\rangle\preceq\langle E,t\rangle$ if and only if $D\subseteq E$ and $s=t\upharpoonright D$. 
Suppose that $\mathscr{C}$ is a chain in $\mathscr{P}$. Let $D=\bigcup_{\langle C,s_C\rangle\in\mathscr{C}}C\subseteq S$, and let $t=\bigcup_{\langle C,s_C\rangle\in\mathscr{C}}s_C$; the definition of $\preceq$ ensures that $t:C\to S$. Clearly $t\upharpoonright C=s_C$ for each $\langle C,s_C\rangle\in\mathscr{C}$.

You may not be used to working with functions as sets of ordered pairs in this way; if so, you can define $t$ in the following equivalent way. For each $x\in D$ there is a $\langle C,s_C\rangle\in\mathscr{C}$ such that $\alpha\in C$; set $t(\alpha)=s_C(\alpha)$. To see that $t(\alpha)$ is well-defined, suppose that $\alpha\in C'$ for some other $\langle C',s_{C'}\rangle\in\mathscr{C}$. Then either $\langle C,s_C\rangle\preceq\langle C',s_{C'}\rangle$, in which case $$t(\alpha)=s_C(\alpha)=(s_{C'}\upharpoonright C)(\alpha)=s_{C'}(\alpha)\;,$$ or $\langle C',s_{C'}\rangle\preceq\langle C,s_C\rangle$, in which case $$s_{C'}(\alpha)=(s_C\upharpoonright C')(\alpha)=s_C(\alpha)=t(\alpha)\;.$$ Again it’s clear that $t\upharpoonright C=s_C$ for each $\langle C,s_C\rangle\in\mathscr{C}$.

Now let $F\subseteq A$ be finite. There is a $\langle C,s_C\rangle\in\mathscr{P}$ such that $D\cap F\subseteq C$, and $s_C$ is good, so $$t\upharpoonright(D\cap F)=s_C\upharpoonright(D\cap F)=s_C\upharpoonright(C\cap F)$$ is a transversal for $\mathscr{S}_{C\cap F}=\mathscr{S}_{D\cap F}$. Thus, $t$ is good, and $\langle D,t\rangle\in\mathscr{P}$. Clearly $\langle C,s_C\rangle\preceq\langle D,t\rangle$ for each $\langle C,s_C\rangle\in\mathscr{C}$, so every chain in $\mathscr{P}$ has an upper bound in $\mathscr{P}$. By Zorn’s lemma $\mathscr{P}$ has a maximal element $\langle M,t\rangle$; to complete the proof we need only show that $M=A$. 
If not, let $\alpha\in A\setminus M$; the maximality of $M$ implies that $\mathscr{S}_{M\cup\{\alpha\}}$ has no transversal. For each $x\in S_\alpha$ let $t_x:M\cup\{\alpha\}\to S$ be defined by
$$t_x(\beta)=\begin{cases}
x,&\text{if }\beta=\alpha\\
t(\beta),&\text{if }\beta\in M\;.
\end{cases}$$
Each of the functions $t_x$ is a choice function for $\mathscr{S}_{M\cup\{\alpha\}}$, but none is a transversal, so none of them is injective. Thus, for each $x\in S_\alpha$ there is a $\beta_x\in M$ such that $t(\beta_x)=x$. Let $B=\{\beta_x:x\in S_\alpha\}$, and let $F=B\cup\{\alpha\}$; $F$ is finite, and $t$ is good, so $t\upharpoonright(M\cap F)=t\upharpoonright B$ is a transversal for $\mathscr{S}_B$. By hypothesis $\mathscr{S}_F$ has a transversal $r:F\to S$; let $x=r(\alpha)\in S_\alpha$. Then $\beta_x\in B$, and $$r(\beta_x)\ne r(\alpha)=x=t(\beta_x)\;.$$ 
In other words, if $r$ is any transversal for $\mathscr{S}_{B\cup\{\alpha\}}$, then $r\upharpoonright B\ne t\upharpoonright B$, contradicting the goodness of $t$. It follows that $M=A$ and hence that $t$ is a transversal for $\mathscr{S}$. 
