Proof of an Infinite Binary Sequence
Here is a formal or informal proof, depending on how the probability distribution is defined. 

Let $S_k=\left\{vvs\in\{0,1\}^\infty\mid v\in\{0,1\}^k,\ s\in\{0,1\}^\infty \right\}$, the sequences of infinite binary digits that repeat their initial $k$ binary digits immediately once (or more), where $k\ge1$. 

$$\begin{align}
\text{Prob[$S_k$]}
&amp;\le\frac1{2^{2k}}\#\left\{w\in\{0,1\}^{2k}\mid w\text{ can be extended to an element in } S_k\right\}\\
&amp;\le\frac1{2^{2k}}\#\left\{w\in\{0,1\}^{2k}\mid w=vv \text{ for some } k \text{ binary digits } v\right\}\\
&amp;=\frac1{2^{2k}}\#\left\{v\in\{0,1\}^{k}\right\}\\
&amp;=\frac1{2^{2k}}2^{k}\\
&amp;=\frac1{2^k}\,.
\end{align}$$

Given an integer $i\ge1$ and a prefix-repetitive toss $S$, we know $S$ must be contained in $S_k$ for some $k\ge i$ since there infinitely many initial segments $v$ such that $vv$ is a prefix of S. Hence,
$$\begin{align}
\text{Prob[$S$ is prefix-repetitive]} 
&amp;\le\sum_{k=i}^\infty\text{Prob[$S_k$]}\\
&amp;\le\sum_{k=i}^\infty\frac1{2^{k}}\\
&amp;=\frac1{2^{i-1}}\,.\\
\end{align}$$

Let $i$ go to infinity, we see that 
$$\text{Prob[$S$ is prefix-repetitive]} = 0\,.$$



Here are three related exercises.

Exercise 1. Show that the number of all prefix-repetitive tosses are more than countably infinite. Hence the approach mentioned in the question is unlikely to succeed.

Exercise 2. Show that if the bits of a sequence $S \in \{0, 1\}^\infty$ are chosen by independent tosses of an unfair coin, then Prob[S is prefix-repetitive] = 0.

Exercise 3. An infinite binary sequence $S \in \{0, 1\}^\infty$ is prefix-palindrome if there are infinitely many strings $v \in \{0, 1\}^*$ such that $vv^R$ is a prefix of $S$, where $v^R$ is the reverse of $v$. Show that if the bits of a sequence $S \in \{0, 1\}^\infty$ are chosen by independent tosses of a coin, then Prob[S is prefix-palindrome] = 0.