Find uncountable set of functions from countable set agreeing only in finite subsets (Sorry for the terrible title, but I couldn't come up with a reasonable description of the problem with words.)
The problem:
Let $I$ be a countably infinite set and $A$ be an infinite set.  


*

*Prove that there exist $S \subseteq A^I$ such that $|S|=\aleph_1$ and for $f,g \in S$, if $f \neq g$ then $\{ i \in I : f(i) = g(i) \}$ is finite.  

*Show that, if $U$ is a free ultrafilter over $I$, then $|A^I/U| \geq \aleph_1$. (Here, the quotient is over the equivalence relation $\sim_U$ given by $a \sim_U b$ iff $\{ i \in I : a(i) = b(i) \} \in U$).
I am allowed to use the Continuum Hypothesis.

Partial resolution:
Given the result of the first part, the second one is easy to prove.
If $U$ is a free ultrafilter over $I$, then all the cofinite sets of $I$ belong to $U$, and therefore, no finite subset of $I$ belongs to $U$.
So, $a/U \neq b/U$ whenever $\{ i \in I : a(i) = b(i)\}$ is finite, and thus, each member of $S$ represents a different class of $\sim_{U}$, that is, a different element of $A^{I}/U$, so that $|A^I/U| \geq |S| = \aleph_1$.
Tentative of solving the first part:
If $A$ is uncountable, then $|A| \geq \aleph_1$ and so we can take a subset $B$ of $A$ with $|B| = \aleph_1$ and the $S$ to be the constant functions from $I$ to $B$.
Suppose $A$ is countable.
In this case, I though I could parametrize a family of functions from $I$ to $A$ more or less like $c_{a,J}$, for $a \in A$ and $J$ a finite subset of $I$, by making
$$c_{a,J}(i) = 
\begin{cases}
a &\text{ if } i \in J,\\
\phi(i) &\text{ if } i \notin J.
\end{cases}$$
where $\phi$ would be a function from $I$ to $A$ that would suit the required conditions.
(The reason for which $J$ must be finite is that otherwise we could have $J,J'$ for the same $a \in A$, so that $c_{a,J}$ and $c_{a,J'}$ could agree in an infinite subset of $I$, although being different.)
But on a second though I realized that this family is not large enough, because, since there are only as many finite subsets of $I$ as members of $I$, there are only $|I| \cdot |A|$ such functions, that is, we would have $|S| = \aleph_0$.
 A: (Let me start by thanking the user martin.koeberl who gave a suggestion in a comment, based on which I came up with this solution.)
We may, without loss of generality, suppose that $I = \mathbb{N}$.
As it was noted in the question, the result is straightforward if $A$ is uncountable (just take the constant maps), so we may suppose that $A = \mathbb{N}$ too (or we could invoke Lowenheim—Skolem Theorem and claim that $A$ actually has a monoid structure, using additive notation, with $+$ and $0$).
So let us suppose, for a contradiction that we have set of functions $S = \{ f_{n} : n \in \mathbb{N} \}$, that is countable and maximal with respect to $f,g \in S$ and $f \neq g$ implies that $\{ n \in \mathbb{N} : f(n) = g(n) \}$ is finite.
Let us define $f : \mathbb{N} \to \mathbb{N}$ by
$$f(n) = \sum_{i=1}^{n} f_{i}(n).$$
We have $f(n) = f_{j}(n)$ iff $f_{i}(n) = 0$ whenever $i<j$.
Thus,
$$\{ n \in \mathbb{N} : f(n) = f_{j}(n) \} \subseteq \bigcap_{i,k=1}^{j-1} \{ n \in \mathbb{N} : f_{i}(n) = f_{k}(n) \},$$
which is finite.
In particular, it follows that $f \notin S$ (if $f = f_{j}$, then $\{ n \in \mathbb{N} : f(n) = f_{j}(n) \} = \mathbb{N}$).
So we may conclude that no countable set of such functions is maximal, whence there is an uncountable such set.
