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Indeed, if $X$ is an infinite set and $I$ has cardinality greater than that of $X^{\aleph_0}$ then $X$ can't contain $I$ distinct subsets with pairwise finite intersection. This answers your question since $c^{\aleph_0}=c$.

Indeed, let $(A_i)_{i\in I}$ be a family of subsets of $X$ with pairwise finite intersection. Let $B_i$ be the set of infinite countable subsets of $X$ contained in $A_i$. Then the $B_i$ are pairwise disjoint. Moreover, $B_i$ is empty only when $A_i$ is finite, and we can remove such exceptional $i$'s because the number of finite subsets of $X$ is only the cardinality of $X$.

The $B_i$ live in the set of infinite countable subsets of $X$, which has cardinality $X^{\aleph_0}$. So $I$ is at most the cardinal of $X^{\aleph_0}$.

Edit: the obvious generalization of the argument is the following: if $\alpha,\beta,\gamma$ are infinite cardinals, and if $\alpha$ admits $\beta$ subsets with pairwise intersection of cardinal $<\gamma$, then $\beta\le\alpha^\gamma$. In particular, if $\alpha=2^\delta$ and $\gamma\le\delta$ then $\alpha^\gamma=\alpha$, so the conclusion reads as: $2^\delta$ does not admit more that $2^\delta$ subsets with pairwise intersection of cardinal $<\delta$.

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Indeed, if $X$ is an infinite set and $I$ has cardinality greater than that of $X^{\aleph_0}$ then $X$ can't contain $I$ distinct subsets with pairwise finite intersection. This answers your question since $c^{\aleph_0}=c$.
Indeed, let $(A_i)_{i\in I}$ be a family of subsets of $X$ with pairwise finite intersection. Let $B_i$ be the set of infinite countable subsets of $X$ contained in $A_i$. Then the $B_i$ are pairwise disjoint. Moreover, $B_i$ is empty only when $A_i$ is finite, and we can remove such exceptional $i$'s because the number of finite subsets of $X$ is only the cardinality of $X$.
The $B_i$ live in the set of infinite countable subsets of $X$, which has cardinality $X^{\aleph_0}$. So $I$ is at most the cardinal of $X^{\aleph_0}$.