TITLE: Does ZF+AD settle the original Suslin hypothesis?
QUESTION [25 upvotes]: Everyone knows that the real line $\langle\mathbb{R},<\rangle$ is
the unique endless complete dense linear order with a countable
dense set. Suslin's
hypothesis is
the question whether we can replace separability in this
characterization with the assertion that the order has the
countable chain condition, that is, that every set of disjoint
intervals is countable. In other words, Suslin's hypothesis, in the
original formulation, is the assertion that the real line is the
unique endless complete dense linear order with the countable chain
condition.
Question. Does ZF plus the axiom of determinacy AD imply the original Suslin hypothesis?
Set theorists proved in ZFC that Suslin's hypothesis is equivalent
to the assertion that there is no Suslin tree, which is a tree of
height $\omega_1$ with no uncountable chains or antichains. And ZF plus the
axiom of determinacy AD settles this version of SH by proving that
$\omega_1$ is measurable and hence weakly compact and hence has the
tree property and so under AD there is there is no
Suslin tree.
So under AD there is no $\omega_1$-Suslin tree. But does this mean that there is no Suslin line? 
The point is that refuting a Suslin tree does not seem directly to refute all Suslin lines in the non-AC context, and therefore it does not seem to settle the original Suslin problem under AD. So the question is: does
ZF+AD settle the original Suslin problem?
Please feel free to post an answer explaining the precise details
of the argument that AD implies there is no $\omega_1$-Suslin tree.
I heard this question this evening at a party at a conference in
honor of Simon Thomas
from a certain prominent set theorist, aged Scotch in hand, who
told me that he would rather not be mentioned, but who said he was
fine for the question to be posted.
Let me update the question (after Asaf's very nice answer) to ask about the situation where we also have the axiom of dependent choice DC. And perhaps one really wants to know about the case in $L(\mathbb{R})$ under AD.
Question. Does ZF+AD+DC imply the original Suslin hypothesis? 
Question. Does $L(\mathbb{R})$ have a Suslin line assuming AD?

REPLY [10 votes]: Nice question! But which aged set theorist drinks prominent Scotch?
The following does not answer this question either. Let $T \in L$ be your favorite Suslin tree of $L$. Consider the $L({\mathbb R})$ of $L[g]$, where $g$ is generic for adding $\omega_1$ Cohen reals. $T$ is still a Suslin tree in $L[g]$. The set of all maximal branches thru $T$ is then a Suslin line in $L({\mathbb R})^{L[g]}$, and $L({\mathbb R})^{L[g]}$ is a model of ZF plus DC plus "there is no w.o. of the reals." So a Suslin line doesn't give a w.o. of ${\mathbb R}$. 
As has been poited out, this approach can't be generalized to show that in the presence of large cardinals (or just ${\sf AD}^{L({\mathbb R})}$), $L({\mathbb R})$ has a Suslin line, as then no $T \in M$, $M$ any inner model of $L({\mathbb R})$ satisfying choice, can be an Aronzsajn tree in $L({\mathbb R})$. 
But, assuming large cardinals, the question if $L({\mathbb R})$ has a Suslin line should have an answer as ${\sf ZFC}$ is $\Omega$-complete concerning truth in $L({\mathbb R})$.