TITLE: Two questions from combinatorics on words
QUESTION [15 upvotes]: Question 1. Assume that an infinite word $u\in\{0,1\}^{\mathbb Z}$ is not balanced. Is it true that there exists a finite 0-1 word $w$ such that $0w01w1$ or $1w10w0$ is a factor of $u$? Is it true that both are? (Perhaps, with different $w$.) - answered in the negative
Question 1 (modified). Assume that an infinite word $u\in\{0,1\}^{\mathbb Z}$ is not balanced. Is it true that there exists a finite (possibly, empty) 0-1 word $w=w_1\dots w_k$ such that $0w_1\dots w_k01w_k\dots w_11$ or $1w_1\dots w_k10w_k\dots w_10$ is a factor of $u$? Is it true that both are? (Perhaps, with different $w$.) - ANSWERED by Wolfgang
Question 2. Assume now that $u\in\{0,1\}^{\mathbb Z}$ is balanced and aperiodic (i.e., Sturmian). Is it true that for any $n\ge1$ there exists a finite 0-1 word $w=w_1\dots w_k$ with $k\ge n$ such that $w_1\dots w_k01w_k\dots w_1$ or $w_1\dots w_k10w_k\dots w_1$ is a factor of $u$? Again, is it true that both are (with different $w$)? - ANSWERED by domotorp
I am familiar with Lothaire's famous monograph "Algebraic Combinatorics on Words" but couldn't find these results there. 
EDIT. Here's what is meant by `balanced'. For a finite 0-1 word $w$ let $\delta(w)$ denote the number of 1s in $w$. A 0-1 word $w$ (finite or infinite) is called balanced if for any $n\ge1$ and any two factors $u$ and $v$ of $w$ of length $n$ we have $|\delta(u)-\delta(v)|\le 1$. 
For instance, $0100101$ is balanced whilst $101000$ is not, since $\delta(101)=2$ and $\delta(000)=0$.

REPLY [3 votes]: Too late, but here's my proof that the answer to Question 1 (modified) is yes.
If $u\in\{0,1\}^\mathbb{Z}$ is not balanced, there exists a palindrome $w$ such that $0w0$ and $1w1$ are factors of $u$ (see Prop. 2.1.3 in Lothaire 2); if $w$ is of minimal length $n$, then $u$ has at most $k+1$ factors of each length $k\leq n+1$ (Prop. 2.1.2).
Moreover $\{0w0,1w1\}$ is the only imbalance among factors of length $\leq n+2$ in $u$. It follows that the set of factors of length $n+2$ is contained in $\{0w0,1w1\}\cup [w01]$, where [x] denotes the conjugacy class of $x$ (all of its cyclic shifts). Also, every factor of length $n+1$ except $0w$ and $1w$ are not right special, i.e., they always occur followed by the same letter.
From this we can derive that there exists $m\geq 0$ such that every occurrence of $0w0$ is followed by $(1w0)^m$ and every occurrence of $1w1$ is followed by $(0w1)^m$. If such $m$ is chosen to be maximal, then either $$0w0(1w0)^m1w1(0w1)^m=0(w01)^mw01w(10w)^m1$$ or
$$1w1(0w1)^m0w0(1w0)^m=1(w10)^mw10w(01w)^m0$$ is a factor of $u$.