Here is my answer to the updated question:

If you allow the forcing $P_\alpha$ to be trivial, then the Cohen real forcing example still works: add a Cohen real, and then perform $\kappa$ many stages of trivial forcing, with countable support. Overall, this is the same as just adding a Cohen real, since the later stages don't do anything, and it will still have the properties you request as in the original question.

But if you insist that every $P_\alpha$ is nontrivial as a forcing notion, then the inaccessibility of $\kappa$ will necessarily be destroyed. To see this, suppose that each $P_\alpha$ has incompatible conditions. Pick a condition $p_\alpha$ in $P_\alpha$ such that there are other conditions in $P_\alpha$ incompatible with $p_\alpha$. Consider any block $B$ of $\omega_1$ many coordinates $\alpha$ at which such nontrivial forcing is to be done. Let $G$ be $V$-generic for the iteration. For each such $\alpha\in B$, let $s_B(\alpha)$ be $0$ or $1$ accordingly as to whether $p_\alpha\in G$ or not. Thus, for each block $B$, the binary pattern of $s_B$ determines a subset of $\omega_1$. Since your iteration uses only countable support, it is dense that the binary pattern of $s_B$ is different from that of $s_{D}$, if $B$ is disjoint from $D$, since we may introduce such a difference by moving beyond the support of any given condition. Since we may divide $\kappa$ into $\kappa$ many disjoint blocks $B$, it follows that in $V[G]$ we have $\kappa$ many subsets of $\omega_1$, and so $\kappa$ is no longer inaccessible, contrary to 1.

Here is my answer to the original question:

Stefan's comment provides an easy example if you don't intend that the forcing should not add bounded sets to $\kappa$. Just add a Cohen real. This of course preserves GCH, has countable conditions, and adds a subset to $\kappa$ (since every subset of $\omega$ is already a subset of $\kappa$); if you throw in all infinite ordinals as well, then you've got a new subset of $\kappa$ of size $\kappa$.

But even if you don't want to add bounded sets to $\kappa$, there are some easy examples showing that the question is probably not quite what you want to ask. This is because the *size* of conditions is not a fundamental feature of a forcing notion. For any partial order $\mathbb{P}$, let's make an isomorphic copy $\mathbb{P}'$ by replacing each condition $p$ with its singleton $p'=\{p\}$. We order $p'\leq q'$ just in case $p\leq q$. So now we have a partial order all of whose conditions have size $1$, and hence are countable. So any forcing notion is isomorphic and hence forcing equivalent to a forcing notion with countable conditions. Thus, this property is probably not really what you want to ask about.