Betting system and upper bounds 
Define $X_0=\alpha\in(0,1)$ the initial capital and $X_n$ as the remaining capital after each game.
A player bets $1-X_n$ if $X_n>1/2$ and $X_n$ if $X_n\leq 1/2$ such that each game is a Bernoulli$(1/2)$.
Define $A_n=\{X_n\in(0,1)\}$, the event that the player neither wins everything or reachs ruin. Show by induction that $P(A_n)\leq 2^{-n}$.

If $\alpha< 1/2$ then the player either goes broken with probability $1/2$ or owns $2X_n$ in the next game. If $\alpha >1/2$ then the player owns every available resource with probability $1/2$ or owns $2X_n -1$ in the next game. Should I use total law of probability and try to work with the conditionals upon the last fortune as in
$$P(A_n) = \sum_{k=1}^m P(A_n|B_m)P(B_m)$$
where $\bigcup B_m= \Omega$.
Also if this is the right path, should I split three-ways with a $X_n=1/2$ case where the player reach either $0$ or $1$?
I'm having a little trouble working this problem out.
 A: UPDATE: This is a new version of the answer. The previous one was very similar but at the end I had in mind a mean over all $\alpha_s$ according to a uniform distribution,  and claiming that equality holds. Now I consider $\alpha$ fixed, that probably is what the OP wants.
We have:
$P(A_n)=P(X_n \in (0,1))=1/2P(0<X_{n-1}<1/2)+1/2P(1/2< X_{n-1}<1)$
The idea here is that if $X_{n-1}=0,1/2,1$ than we are for sure either ruined or full of money at $X_n$. In all other cases we have 1/2 chance of ending up either ruined or full of money.
This means that:
$P(A_n)=1/2P(A_{n-1})-1/2P(X_{n-1}=1/2)\le P(A_{n-1})/2$
Now from here we have the inequality easily by induction.
A: You can't ignore the possibility that $\ X_n=\frac{1}{2}\ $, because it will occur with  probability $\ \frac{1}{2^n}\ $ if $\ \alpha\ $ has a terminating binary expansion $\ \sum_\limits{i=1}^n\frac{\alpha_i}{2^i}+\frac{1}{2^{n+1}}\ $, so you definitely need to include this case in your analysis.  So your initial cases are:

*

*If $\ \alpha<\frac{1}{2}\ $, then the player goes broke with probability $\ \frac{1}{2}\ $ or plays the next game with a capital of $\ 2\alpha\ $, again with probability $\ \frac{1}{2}\ $.  This latter occurrence is the event $\ A_1 $, so $\ P(A_1)=\frac{1}{2}\ $ in this case.

*If $\ \alpha=\frac{1}{2}\ $, then the player goes broke with probability $\ \frac{1}{2}\ $ or wins everything with probability $\ \frac{1}{2}\ $.  Therefore, no second game takes place in this case, and $\ P(A_1)=0\ $.

*If $\ \alpha>\frac{1}{2}\ $ then the player wins everything with probability $\ \frac{1}{2}\ $ or plays the next game with a capital of $\ 2\alpha-1\ $, again with probability $\ \frac{1}{2}\ $. As in the first case, this latter occurrence is the event $\ A_1 $, so $\ P(A_1)=\frac{1}{2}\ $ in this one too.

Thus, in all three cases $\ P(A_1)\le\frac{1}{2}\ $, which gives you the base proposition for a proof by induction.
For the induction step, your suggestion of using the law of total probability is a good one, but you need to choose an appropriate set of $\ B_m\ $.  If $\ n\ $ is a positive integer such that $\ P(A_n)\le\frac{1}{2^n}\ $, I'd suggest taking
\begin{align}
B_1&=\Omega\setminus A_n ,\\
B_2&=\left\{0<X_n<\frac{1}{2}\right\} ,\\
B_3&=\left\{X_n=\frac{1}{2}\right\} ,\\
B_4&=\left\{\frac{1}{2}<X_n<1\right\} .
\end{align}
Then $\ B_i\ $ are mutually exclusive events with $\ \bigcup_\limits{i=2}^4B_i= A_n\ $ and $\ \bigcup_\limits{i=1}^4B_i=\Omega\ $.  By the same argument as in the case $\ n=0\ $ above, we have
\begin{align}
P\left(A_{n+1}\,\left|\,0<X_n<\frac{1}{2}\right.\right)&=\frac{1}{2}\ ,\\
P\left(A_{n+1}\,\left|\,X_n=\frac{1}{2}\right.\right)&=0\ \text{, and}\\
P\left(A_{n+1}\,\left|\,\frac{1}{2}<X_n<1\right.\right)&=\frac{1}{2}\ ,
\end{align}
and since $\ A_{n+1}\subseteq A_n\ $ (i.e. $\ A_{n+1}\ $ can't occur unless $\ A_n\ $ has also occurred), we also have
$$
P\left(A_{n+1}\,\left|\,\Omega\setminus A_n\right.\right)=0\ .
$$
The law of total probability therefore gives us
\begin{align}
P(A_{n+1})&=\sum_{i=1}^4P(A_{n+1}\,|\,B_i\,)P(B_i)\\
&=P\left(A_{n+1}\,\left|\,0<X_n<\frac{1}{2}\right.\right)P\left(\,0<X_n<\frac{1}{2}\right)\\
&\hspace{1.5em}+P\left(A_{n+1}\,\left|\,\frac{1}{2}<X_n<1\right.\right)P\left(\,\frac{1}{2}<X_n<1\right)\\
&=\frac{1}{2}\left(P\left(\,0<X_n<\frac{1}{2}\right)+P\left(\,\frac{1}{2}<X_n<1\right)\right)\\
&\le\frac{1}{2}P(A_n)\ ,
\end{align}
because $\ \left\{\,0<X_n<\frac{1}{2}\right\}\cup\left\{\,\,\frac{1}{2}<X_n<1\right\}\subseteq A_n\ $. From the induction hypothesis, $\ P(A_n)\le\frac{1}{2^n}\ $, it therefore follows that
$$
P(A_{n+1})\le \frac{1}{2^{n+1}}\ ,
$$
which completes the proof by induction.
