The following (negative answer to Erdos' question) will appear in a joint work with Shelah.
Claim: Suppose $V \models 2^{\aleph_0} = \lambda > \kappa = \aleph_1$. Let $P$ add $\kappa$ Cohen reals. Then in $V^{P}$, there is no such family.
Proof: Let $r \in {}^{\kappa}2$ be the Cohen generic sequence. Clearly $V[r] \models 2^{\aleph_0} = \lambda$. Suppose $\langle f_{\alpha} : \alpha < \lambda \rangle$ is a sequence of pairwise distinct analytic functions in $V[r]$. Choose $X \in [\lambda]^{\lambda}$, $\xi_{\star} < \kappa$ such that for each $\alpha \in X$, $f_{\alpha}$ is coded in $V[r \upharpoonright \xi_{\star}]$. Let $z_{\star} \in \mathbb{C}$ be Cohen over $V[r \upharpoonright \xi_{\star}]$. Since two distinct analytic functions only agree on a countable set, it follows that $\langle f_{\alpha}(z_{\star}) : \alpha \in X \rangle$ are pairwise distinct.
Update: Shelah and I showed that the answer to Erdos' question is independent of ZFC + not CH so the other direction also holds.
An interesting question that remains open is: Is the following consistent: $2^{\aleph_0} = \aleph_2$ and there exists $U \in [\mathbb{C}]^{\aleph_1}$ such that for every $A \in [\mathbb{C}]^{\aleph_1}$, there is a non constant entire map that sends $A$ into $U$?