What can we say about the one point compactification of a Suslin tree? As a continuation to this question:

The Alexandroff one point compactification of $(X,\tau)$, is a space $X \cup \{a\}$, where open neighborhoods of $X$ are $\{ U : U \in \tau\} \cup \{ V \cup \{a\}: V \in \tau $ and $V^C$ is closed and compact in $ X \}$
A tree $T$ is a Suslin tree if: 
  1. The height of $T$ is $\omega_1$.
  2. Every branch in $T$ is at most countable.
  3. Every antichain in $T$ is at most countable.

Let $X=T \cup \{q\}$ be the Alexandroff one point compactification of a Suslin tree. 
I am trying to figure out whether $X$ is Frechet-Urysohn, and whether ONE wins in $G_{np}(q,X)$.
Any ideas or directions?
Thank you! 

From this answer we have that if $X = \{ \infty \} \cup T$ is the one-point compactification of an Aronszajn tree $T$ with the tree topology, then the sets of the form $$U ( s_1 , \ldots , s_n ) = \{ \infty \} \cup {\textstyle \bigcap_{i \leq n}} \{ x \in T : x \not\leq_T s_i \}$$ where $s_1 , \ldots , s_n \in T$ form a neighbourhood basis at $\infty$.
The basic idea for show that $X$ is Fréchet-Urysohn is to notice first (and Henno Brandsma has) that for each $s \in T$ we have  $\chi ( X , s ) \leq \aleph_0$, and so $X$ is "Fréchet-Urysohn at $s$."
Given $A \subseteq T$, it is relatively easy to show that $\infty \in \overline{A}$ iff either 


*

*$A$ includes an infinite family of pairwise incomparable nodes of $T$; or

*$A$ includes an infinite chain with no upper bound in $T$.


(If $U = U(s_1 ,\ldots , s_n )$ is a basic open neighbourhood of $\infty$ and $A$ satisfies either (1) or (2), then there must be points of the antichain/chain which are not below any $s_i$.  If $A$ fails to satisfy either of these conditions, then $A$ is the union of finitely many chains, each of which has an upper bound in $T$, and so $U(s_1 , \ldots , s_n)$ — where the $s_i$ are such chosen upper bounds — is an open neighbourhood of $\infty$ disjoint from $A$.)
In each of the above cases you basically do the obvious thing to construct a sequence in $A$ converging to $\infty$.
(That there is no winning strategy for ONE in $G_\text{np} ( \infty , X )$ is in the linked answer.)
