Logic: Infinite $K$-sequences, and $n$-free problems There are two problems which seem to be similar that I am stuck on.  This is not homework for me but would very much like to understand what is going on.
First problem is this:
  An infinite $k$-sequence is a function $s: \Bbb{N} \to \{0, \ldots ,k-1\}$ and a $k$-sequence of length $l$ is a function $s: \{0,\ldots ,l-1\} \to \{0, \ldots ,k-1\}$.
We say that a $k$-sequence $s$ (finite or infinite) is $n$-free if there does not exist a finite $k$-sequence $x$ such that $x^n$ (repeated $n$ times) is a substring of $s$.

a.  Show that there is no infinite $2$-sequence that is $2$-free.
b. Using Konig's lemma, show that there is an infinite $3$-free $2$-sequence if and only if for every $n$, there is a $3$-free $2$-sequence of length $n$.

Attempt at a:
A $2$-sequence is one s.t. s: $\Bbb{N} \to \{0,1\}$ (there are an infinite number of ways to map $\Bbb{N} \to \{0,1\}$, e.g. $\{751,752\} \to \{0,1\}$, or I suppose even $\{751,752,753,754\} \to \{0,1\}$ as we could have $\Bbb{N}$ mod $2$. Ok, now if there was an infinite $2$-sequence which was $2$-free then there would be a finite string that repeated $n$ times would be a substring of $s$.  This would give some $\{0,1\}\{0,1\}\{0,1\}\{0,1\}\{0,1\}\ldots\{0,1\} - n$ times, and so for something that is $2$-free we would have either $00, 01, 10,$ or $11$ being an element of $s$. But since $s$ does not contain any double digit elements, there is no $2$-sequence which is $2$-free. 
 - Also, I am wondering, if I am even on the right track, about whether $\{0,1\}\{0,1\}$ would be those elements $\{00,01,10,11\}$.
b. Konig's lemma states that we have a graph $G$ which has an infinite number of vertices, but that each vertice has only a finite adjacency to other vertices, then there exists an infinitely long simple path.  Now if we have a $3$-free $2$ sequence then?  I am a bit confused as above this would not be possible (for the $2$-sequence being $2$-free).  Or perhaps this problem is supposed to be such that, if this $3$-free $2$-sequence could exist then we would have that for every $n$ that there is a $3$-free $2$-sequence of length $n$ and vice versa?  both of them however may not actually exist, but that given one, the other exists?
Thanks.
 A: I don't understand your attempt at (a), because you write $\mathbb N\to\{0,1\}$ but then give maps whose domain is much less than $\mathbb N$, like $\{751,752\}$. Also, the sentence beginning "OK, now $\dots$" refers to a 2-free sequence but then essentially quotes the definition of "not $n$-free".  
I think (a) becomes much easier if you don't worry about $n$-freeness for any $n$ other than 2, and just see what the options are for the first few terms in a (hypothetical) 2-free 2-sequence.  Note that "2-free" just means that no finite substring can be immediately repeated.  So, after you choose either 0 or 1 as the first term of your hypothetical infinite sequence, all subsequent terms are forced, since you can't immediately repeat even a single term.  But then you'll find an immediate repetition of a two-term subsequence, already in the first four terms.
For the non-trivial direction of (b), form a graph $G$ whose nodes are the finite 3-free 2-sequences (of all lengths), and put an edge joining two such sequences $u$ and $v$ if one of them is obtained from the other by adding a single term (0 or 1) at the end.  Use König's Lemma to get an infinite path $P$ in $G$, let $v$ be a vertex on this path whose length (as a 3-free 2-sequence) is as short as possible, and contemplate the part of $P$ after $v$.
