Does adding $\kappa$ Cohen reals preserve a Suslin tree? I have tried to prove the following problem in Jech (Exercise 16.14):

For a Suslin tree $T$ and a forcing $P=\operatorname{Col}(\kappa, 2)$ adjoins $\kappa$ Cohen generic reals, show that $T$ is a Suslin tree in $V[G]$ for a generic filter $G$ over $P$.

Jech gives a hint on how to show that $T$ is c.c.c. in $V[G]$, and checking the height of $T$ is easy. However, I feel that it is not an end of a proof. I need to show that every chain of $T$ is countable.
The problem is that $P$ may add an uncountable branch. Can we rule out such a case? Since $T$ itself is a c.c.c. forcing that adds an uncountable branch of $T$, just a countable chain condition is not sufficient to rule out that case. I would appreciate your help.

From an uncountable chain $C$ in $T$, you can build an uncountable antichain! For each node $\sigma\in C$, consider some node $\tau_\sigma\in T$ which is above $\sigma$ but not on $C$ (so "splits off" from $T$). Any two such nodes are incompatible, and there are uncountably many of them.
In case $T$ has no dead ends (which we can assume WLOG), this $\tau_\sigma$ can be an immediate successor of $\sigma$, in which case the map $\sigma\mapsto\tau_\sigma$ is injective; this makes it trivial to check that $\{\tau_\sigma: \sigma\in C\}$ is uncountable, but is not necessary. (HINT: if $\{\tau_\sigma:\sigma\in C\}$ were countable, then it would contain only nodes of rank bounded by some $\alpha<\omega_1$ . . .)
