$\omega$-colorings of $\kappa^2$ Let $\kappa\le 2^{\aleph_0}$ be an infinite cardinal. We have a collection of functions $\{f_i|i<\kappa\}$ such that $f_i:i\rightarrow \omega$ and the collection is "triangle-free", i.e. there are not $i<j<k<\kappa$ such that $$f_j(i)=f_k(i)=f_k(j).$$
Is it always possible to extend this collection by adding one more function $f_\kappa:\kappa\rightarrow\omega$ so that the collection $\{f_i|i\le\kappa\}$ remains triangle-free?
 A: No, it is not always possible.
I find it helpful to translate the problem slightly. Note that, if $\alpha$ is an ordinal (not necessarily $\leq 2^{\aleph_0}$), a sequence of functions $\langle f_i \mid i < \alpha \rangle$ as specified is the same as a single function $f:[\alpha]^2 \rightarrow \omega$ (where $[\alpha]^2$ is the set of 2-element subsets of $\alpha$), and the requirement that $\langle f_i \mid i < \alpha \rangle$ is triangle-free precisely the same as requiring $f$ to be triangle-free (i.e. there is no $i < j < k < \alpha$ such that $f(\{i,j\}) = f(\{i,k\}) = f(\{j,k\})$).
Suppose that, for every $\alpha < (2^{\aleph_0})^+$ and every triangle-free coloring $f:[\alpha]^2 \rightarrow \omega$, $f$ can be extended to a triangle-free coloring $g:[\alpha+1]^2 \rightarrow \omega$ with $g \restriction [\alpha]^2 = f$. Then we can recursively build a sequence $\langle g_\alpha \mid \alpha < (2^{\aleph_0})^+ \rangle$ such that $g_\alpha:[\alpha]^2 \rightarrow \omega$ and, for $\alpha < \beta$, $g_\beta$ extends $g_\alpha$. Then, letting $g = \bigcup_{\alpha < (2^{\aleph_0})^+}g_\alpha$, we have that $g:[(2^{\aleph_0})^+]^2 \rightarrow \omega$ is a triangle-free coloring, contradicting the Erdos-Rado theorem. There thus must be some $\alpha < (2^{\aleph_0})^+$ and $f:[\alpha]^2 \rightarrow \omega$ such that $f$ is triangle-free and cannot be extended to a triangle-free coloring with domain $[\alpha+1]^2$. Taking a bijection between $\alpha$ and its cardinality, we can assume $f$ is actually a triangle-free coloring $f:[\kappa]^2 \rightarrow \omega$ that cannot be extended to a triangle-free coloring on $[\kappa+1]^2$, where $\kappa \leq 2^{\aleph_0}$ is a cardinal. Translating back to your formulation, we get $\langle f_i \mid i < \kappa \rangle$ by letting, for $i < j < \kappa$, $f_j(i) = f(\{i,j\})$. This sequence cannot be extended by a function $f_\kappa$.
As a side note, there is a very natural triangle-free coloring of size continuum that cannot be extended. Let ${^\omega}2$ denote the set of functions $F:\omega \rightarrow 2$, and define $f:[{^\omega}2]^2 \rightarrow \omega$ by letting $f(F,G)$ be the least $n$ such that $F(n) \neq G(n)$. $f$ is obviously triangle-free, and there's a nice argument showing that it cannot be extended. 
