Proof that the Axiom of Continuity is unique. First I will outline the notation that I am going to use. A preference relation of player $i$ over a set of outcomes $O$ is a binary relation denoted by $\succeq_i$.
$x \succeq_i y$ is read "player $i$ prefers $x$ to $y$ or is indifferent between the two outcomes".
$x \succ_i y$ is read "player $i$ prefers $x$ to $y$.
$ x \approx_i y$ is read "player $i$ is indifferent between the two outcomes".
A lottery $L$ in which outcome $A_k$ has probability $p_k$ is denoted by $L = [p_1(A_1),...,p_K(A_K)]$
Now a pair of axioms:
Continuity

For every triplet of outcomes $A \succeq_i B  \succeq_i C$ there exists a number $\theta_i \in [0,1]$ such that $$B \approx_i [\theta_i(A), (1-\theta_i)(C)].$$

Monotonicity

Let $\alpha, \beta$ be numbers in $[0,1]$, and suppose that $A \succ_i B$. Then
$$\alpha \geq \beta \iff [\alpha(A), (1-\alpha)(B)] \succeq_i [\beta(A), (1-\beta)(B)]$$

The theorem I need to prove is

If the preference relation $\succeq_i$ is transitive and satisfies the above axioms, and if $A \succeq_i B  \succeq_i C$, and $A \succ_i C$, then the value of $\theta_i$ in continuity is unique.

Since $A \succeq_i B  \succeq_i C$, by continuity there exists a number $\theta_i \in [0,1]$ such that $B \approx_i [\theta_i(A), (1-\theta_i)(C)]$. Suppose this number is not unique. That is, there exists $\hat{\theta_i} \in [0,1]$ such that $B \approx_i [\hat{\theta_i}(A), (1-\hat{\theta_i})(C)]$ and $\theta_i > \hat{\theta_i}$ without loss of generality.
Since $A \succ_i C$ then by monotonicity
$$\theta_i > \hat{\theta_i}\iff [\theta_i(A), (1-\theta_i)(C)] \succ_i [\hat{\theta_i}(A), (1-\hat{\theta_i})(C)] \iff B \succ_i B$$
Which is a contradiction.
I think I have used monotonicity incorrectly. Can I have some direction on how to prove this correctly? Or In the event that it is correct can I have some comments on my (in my opinion) incorrect usage of the monotonicity axiom?
 A: Since there is no need to mention multiple players, I will drop the subscript $i$.
You have provided us with no axiom that allows us to conclude $A \succ B$. It is perfectly consistent with the axioms you have stated that it is never the case that $A \succ B$, regardless of the behaviour of $\approx$ and $\succeq$. There are also plenty of other bizarre pathologies that could occur in the absence of relationships between $\succ$ and $\succeq$. Therefore, I do not see how you can conclude that $\theta_i > \hat{\theta}_i \implies [\theta_i(A), (1 - \theta_i)(C)] \succ [\hat{\theta}_i(A), (1 - \hat{\theta}_i)(C)]$. You can conclude that $\theta_i > \hat{\theta_i}$ means both that $[\theta_i(A), (1 - \theta_i)(C)] \succeq [\hat{\theta}_i(A), (1 - \hat{\theta}_i)(C)]$ and that it is not the case that $[\hat{\theta}_i(A), (1 - \hat{\theta}_i)(C)] \succeq [\theta_i(A), (1 - \theta_i)(C)]$.
In order to properly solve this problem, we require some additional axioms regarding the behaviour of $\succeq$ and of $\approx$. We now require the following claim, which you have not stated:

Claim: for all $A$, $B$, if $A \approx B$, then $A \succeq B$ and $B \succeq A$.

Now suppose we have $A \succeq B \succeq C$ and $A \succ C$. Suppose we had $\alpha$, $\beta \in [0, 1]$ such that $B \approx [\alpha(A), (1 - \alpha)(C)]$ and $B \approx [\beta(A), (1 - \beta)(C)]$. Then by our claim, we have $[\beta(A), (1 - \beta)(C)] \succeq B \succeq [\alpha(A), (1 - \alpha)(C)]$. By transitivity, we have $[\beta(A), (1 - \beta)(C)] \succeq [\alpha(A), (1 - \alpha)(C)]$. By monotonicity, we have $\beta \geq \alpha$. Repeat the above argument with $\alpha$ and $\beta$ switched to see that $\alpha \geq \beta$. Therefore, $\alpha = \beta$.
