Use compactness theorem to prove existence of a Flip-Set Let $\{0,1\}^w$ be the set of infinite words $a_0a_1...$ with $a_i \in \{0,1\}$ for all $i \in \mathbb{N}$.
A set $F \subseteq \{0,1\}^w$ is called a Flip-Set, if for all words $a,b \in \{0,1\}^w$ that differ in exactly one spot, it holds that $a \in F \iff b \notin F$.
How can I prove the existence of a Flip-Set with the compactness theorem?
 A: General idea: The Compactness Theorem is about concluding the satisfiability of a set of formulas by showing that every finite subset of these formulas is satisfiable. To use this for proving the existence of a flip set, we have to define a set of formulas such that a satisfying interpretation of these formulas can be translated into flip set. We then have to show that every finite subset of these formulas can be satisfied.
Define a variable for every word, $\mathcal V := \{ X_a \mid a \in \{0,1\}^\omega\}$. Given these variables, define a formula for every pair of words differing in one spot, expressing that exactly one of the variables corresponding to these words should be true:
$$\Phi := \{ X_a \leftrightarrow \lnot X_b \mid a ,b \in \{0,1\}^\omega, a \text{ and } b \text{ differ in exactly one position }\}$$
If we can show that $\Phi$ is satisfiable, then we can construct a flip set, containing all words for which the variable for the word is set to true. To show that $\Phi$ is satisfiable, using the Compactness Theorem, we have to show that every finite subset of $\Phi$ is satisfiable.
Consider a finite subset $\Phi_0 \subseteq \Phi$.
Define the equivalence relation $\sim$ on $\{0,1\}^\omega$ as follows. For every $a,b \in \{0,1\}^\omega$, let $a \sim b :\Leftrightarrow a,b$ differ only in finitely many spots (it should be easy to verify this is a reflexive, symmetric and transitive). Lift $\sim$ to an equivalence relation on $\mathcal V$, such that for all $a,b\in \{0,1\}^\omega$, it holds that $X_a \sim X_b :\Leftrightarrow a \sim b$.
Let $\mathcal V(\Phi_0)$ be the set of variables in $\Phi_0$. Let $V_0,\dots,V_{n-1}$ be the $\sim$ equivalence classes of $\mathcal V(\Phi_0)$. This is well-defined, since $\Phi_0$ is finite, so there are only finitely many equivalence classes. Intuitively, $V_0,\dots,V_{n-1}$ define a partition of $\mathcal V(\Phi_0)$ into clusters of words that are similar, such that they can be transformed into each other by flipping finitely many positions. 
For every $i \in \{0,...,n-1\}$, fix some representative $a_i \in V_i$. Define the interpretation $\mathcal I: \mathcal V(\Phi_0) \to \{0,1\}$ such that for $X_a \in \mathcal V(\Phi_0)$ and the unique $i$ with $a \sim a_i$: $$\mathcal I(X_a) := \begin{cases}
1 & a \text{ and } a_i \text{ differ in an even number of positions } \\
0 & a \text{ and } a_i \text{ differ in an odd number of positions } 
\end{cases}$$
This is well-defined because each $V_i$ is finite, by finiteness of $\Phi_0$.
Intuition: We can understand this as a two-coloring of the words in each cluster: We start by coloring the representatives $a_i$ in e.g. green ("true"). Then we color all conflicting words (differing with $a_i$ in exactly one position) in e.g. red ("false"). The words conflicting with those words by differing in one position then differ again with the representative $a_i$ in an even number of positions (zero or two positions). Thus these can be colored green ("true") again. And so on. The initial choice of the representative thus fully determines the coloring of the entire cluster.
Claim: $\mathcal I$ satisfies $\Phi_0$. For every $a,b \in \{0,1\}^\omega$, such that $X_a \leftrightarrow \lnot X_b \in \Phi_0$, consider the unique $i \in \{0,...,n{-}1\}$ with $a \sim a_i$. Since $a,b$ differ in exactly one spot, it follows that also $b \sim a_i$ (both are in the same cluster) and that exactly one of $a,b$ differs with $a_i$ in an even number of position (if one differs in a even number of positions from $a_i$ the other has to differ in an odd number of positions from $a_i$ and vice versa). It follows $\mathcal I(X_a) \neq \mathcal I(X_b)$, and therefore $\mathcal I \models X_a \leftrightarrow \lnot X_b$. Thus $\mathcal I \models \Phi_0$.
We have shown that every finite $\Phi_0 \subseteq \Phi$ is satisfiable. From the Compactness Theorem, it follows that $\Phi$ is satisfiable. Therefore, there is $\mathcal I \models \Phi$. Define the set $F \subseteq \{0,1\}^\omega$ such that $a \in F :\Leftrightarrow \mathcal I(X_a) = 1$. Consider any $a,b \in \{0,1\}^\omega$ that differ in exactly one position. Since $X_a \leftrightarrow \lnot X_b \in \Phi$, it follows $\mathcal I \models X_a \leftrightarrow \lnot X_b$ and therefore $\mathcal I(X_a) \neq \mathcal I(X_b)$. Therefore $a \in F$ iff $b \not \in F$. In conclusion, $F$ is a flip set.
