Largest binary sequence with no more than two repeated subsequences $A$ is an ordered sequence of elements $a_i = 0, 1$ containing no more than two adjacent repeated subsequences $[a_i, a_{i + k})$. What is the longest sequence $A$? Is it even finite?
For example, the subsequence $\{0\}$ is found three times in a row in $\{0, 0, 0\}$, and $\{1, 0\}$ is found in $\{\ldots, 1, 0, 1, 0, 1, 0, \cdots\}$ at least three times in a row. Therefore, both sequences are invalid.
As a simplification, if the number of allowed repeated subsequences is decreased to one, the sequences with the maximum cardinality under this new restriction are simply $\{0, 1, 0\}$ and $\{1, 0, 1\}$, each containing 3 elements.
 A: As Gerry Myerson noted in the comments, you're probably interested in cube-free binary words.  There are many cube-free infinite binary words (words of infinite length).  In fact, there are an infinite number of them.  Furthermore, there are an uncountably infinite number of them.
A: A simple and unusually interesting example is the Thue-Morse sequence.  Let: 
$$\begin{align}
S_0 & = \langle 0\rangle \\
S_1 & = \langle 0, 1\rangle \\
S_2 & = \langle 0, 1, 1, 0\rangle \\
S_3 & = \langle 0, 1, 1, 0, 1, 0, 0, 1\rangle \\
S_4 & = \langle 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0\rangle \\
&\vdots
\end{align}$$
Where $S_{i+1}$ is formed by concatenating $S_i$ with its own inverse, which has 1 wherever $S_i$ has 0 and vice-versa. The Thue-Morse sequence is the "limit" of this process: The $i$th element of the Thue-Morse sequence is the same as the $i$th element of the first $S_n$ that is long enough to have an $i$th element.
$TM$ satisfies the following recurrence:
$$
\begin{align}
TM(0) & = 0 \\
TM(2n) & = TM(n) \\
TM(2n+1) & = 1-TM(n)
\end{align}
$$
This sequence $TM$ has the property that for any sequence $X$, $TM$ does not contain $XXX$.
