An exercise in infinite combinatorics from Burris and Sankappanavar Exercise 6.7 in chapter IV of Burris and Sankappanavar's A Course in Universal Algebra starts as follows:

Show that for $I$ countably infinite there is a subset $S$ of the set of functions from $I$ to $2$ which has cardinality equal to that of the continuum such that for $f \not= g$ with $f,g \in S$, $\{i \in I : f(i) = g(i)\}$ is finite.

I don't see how $S$ can have cardinality more than 2: let $f, g, h \in S$, with $f \not= g$ and $f \not= h$ and let $A = \{i \in I : f(i) = g(i)\}$ and $B = \{i \in I : f(i) = h(i)\}$. For $i$ in $I\backslash (A \cup B)$ we have $g(i) \not= f(i) \not= h(i)$, whence $g(i) = h(i)$, since $f(i), g(i), h(i) \in 2 = \{0, 1\}$. But $A$ and $B$ are finite, $I \backslash(A\cup B)$ is infinite. So $g(i) = h(i)$ holds for infinitely many $i$ and we must have $g = h$.
What have I missed? (A redefinition of 2, perhaps?)
I would also be grateful for a reference for what appears to be the goal of this exercise, namely to use an ultraproduct construction to obtain uncountable models from countable ones.
 A: On the website of the book there is a newer version and - as the OP already pointed out in their comment the exercise is already corrected there (see page 150). It says:

Show that, for $I$ countably infinite and $A$ infinite, there is a subset $S$ of the set of
  functions from $I$ to $A$ which has cardinality equal to that of the continuum such that for
  $f\ne g$ with $f,g\in S$, $\{i\in I; f(i)=g(i)\}$ is finite. Conclude that $|A^I/U|\ge 2^\omega$ if $U$
  is a nonprincipal ultrafilter over $I$.


The question basically boils down to finding such family for functions from an infinite countable set to an infinite countable set. (Since $A$ is infinite, it contains an infinite countable set.)$\newcommand{\N}{\mathbb{N}}\newcommand{\Q}{\mathbb{Q}}\newcommand{\R}{\mathbb{R}}$
So we can take $I=\N$ and $A$ can be an arbitrary infinite countable set.

For $A=\Q$ we can repeat basically the same trick as the trick which is used to show the existence of an almost disjoint family of cardinality $\mathfrak c$. See, for example, here or here.
For any real number $r\in\R$ we take any injective sequence $f_r \colon \N \to \Q$ of rational numbers, which converges to $r$. The system $\{f_r; r\in\R\}$ has the same cardinality as $r$. And for any $r\ne r'$ the set $\{i\in\N; f_r(i)=f_{r'}(i)\}$ is finite, since these two sequences have different limits.

A different approach, which only shows that there is uncountably many such functions.
Lemma: If $\{f_n; n\in\N\}$ is a sequence of functions from $\N$ to $\N$ then there is a function $f$ such that each set $A_n=\{i\in\N; f(i)=f_n(i)\}$ is finite.
Proof. We can define
$$f(n)=\max\{f_i(j); i=0,\dots,n-1; j=0,\dots,n-1\}+1.$$
Then we have $\{i\in\N; f_n(i)=f(i)\}\subseteq\{0,1,\dots,n\}$.
$\qquad\square$
Using the above fact and transfinite induction we can get a family of functions with cardinality $\aleph_1$, which has the required properties.
