EDIT: I'll leave my previous answer up for now (at the end of this one), but here's an easier answer that doesn't need assumptions like CH that go beyond ZFC.

It's well-known that there is a family of continuum many infinite subsets of $\omega$ such that every two have finite intersection. List such a family as $\{A_\xi:\xi<\mathfrak c\}$. Also, list all the subsets of $\omega$ as $\{S_\xi:\xi<\mathfrak c\}$. For each $\xi$, let $C_\xi$ be an infinite subset of $A_\xi$ that is either included in $S_\xi$ or disjoint from $S_\xi$. Then $\mathcal C=\{C_\xi:\xi<\mathfrak c\}$ serves as the required counterexample.

PREVIOUS ANSWER:
Here's a counterexample for $\kappa=\aleph_0$ assuming the continuum hypothesis.

Fix a nonprincipal ultrafilter $\mathcal U$ on $\omega$ and, using CH, list its elements in an $\omega_1$-sequence as $B_\xi,\ \xi<\omega_1$. Build an $\omega_1$-sequence of sets $C_\xi$ inductively so that (1) each $C_\xi$ is an infinite subset of $B_\xi$, (2) the intersection $C_\xi\cap C_\eta$ is finite for all $\eta<\xi$, and (3) $C_\xi\notin\mathcal U$. To do this, at stage $\xi$, when you already have the countably many sets $C_\eta$ for $\eta<\xi$, list those sets as $C'_n$ for $n\in\omega$ and inductively choose distinct points $x_n\in B_\xi$ such that $x_n\notin\bigcup_{k<n}C'_k$. These choices can be made because $B_\xi$ is in $\mathcal U$ and each $C'_k\notin\mathcal U$ by induction hypothesis (3). If we were to take $C_\xi=\{x_n:n\in\omega\}$, we'd satisfy requirements (1) and (2) but perhaps not (3). So split this $\{x_n:n\in\omega\}$ into two infinite pieces; at least one of the pieces is not in $\mathcal U$, and we take that piece as $C_\xi$.

The family $\mathcal C=\{C_\xi:\xi<\omega_1\}$ satisfies the cardinality requirements in the question: its members $C_\xi$ are infinite and their pairwise intersections are finite. I claim that it does not have property B. Indeed, for any $S\subseteq\omega$, one of $S$ and $\omega-S$ is in $\mathcal U$, hence is one of the $B_\xi$'s, and hence includes the corresponding $C_\xi$.

Remark: CH is overkill here. Essentially the same argument works if $\mathfrak r$, the unsplitting number, is $\aleph_1$; just enumerate a $\pi$-base for $\mathcal U$ instead of all of $\mathcal U$. Also, $\mathfrak p=\mathfrak u$ seems to be sufficient: Let $\{B_\xi:\xi<\mathfrak u\}$ be a base for $\mathcal U$ (rather than all of $\mathcal U$) and use the assumption about $\mathfrak p$ to obtain the desired $x_n$'s.

I wouldn't be surprised if there's a counterexample in ZFC without any special hypotheses, but I don't yet see one.