Let $x_1$ denote the number picked by player $i$. Let $x_\lt$ and $x_\gt$ be the lesser and greater of $x_1$ and $x_2$. Also, I will use $\delta$ and $\epsilon$ to represent arbitrarily small displacements.
Player $3$ picks either $x_\lt-\epsilon$ or $x_\gt+\epsilon$ or any number (it doesn't matter which) in $(x_<,x_>)$. The payoffs for Player $3$ in these cases are $x_\lt-\epsilon$, $1-x_\gt-\epsilon$ and $(x_\gt-x_\lt)/2$, respectively, and she picks the greatest among these three. Note that in the third case, the half of the interval $[x_\lt,x_\gt]$ that Player $3$ doesn't win goes to Players $1$ and $2$ in equal parts (i.e. they each get one quarter of the interval), since in this case Player $3$ chooses uniformly randomly within that interval.
We can assume without loss of generality that $x_1\le\frac12$.
Now assume first that Player $2$ picks a number above $x_1$. Then we have to distinguish two cases.
For small $x_1$, it will not pay for Player $3$ to use her first option. Then there is a boundary for $x_2$ at which Player $3$ switches form her second to third option. This is the optimal move for Player $2$, since playing to either side of it would just cede territory. The condition for Player $3$ to be indifferent between these two options is $1-x_2=(x_2-x_1)/2$ and thus $x_2=(x_1+2)/3$. At this point $x_2-x_1=(2-2x_1)/3$. If Player $3$ goes for her second option, that interval is split evenly between Players $1$ and $2$, so the payoffs are $((2x_1+1)/3,(1-x_1)/3+\frac\epsilon2,(1-x_1)/3-\frac\epsilon2)$; whereas if Player $3$ goes for her third option, that interval is split in proportions $\frac14:\frac14:\frac24$ (see above), so the payoffs are $((5x_1+1)/6,(1-x_1)/2,(1-x_1)/3)$. At the equilibrium point, the $\epsilon$ difference favours Player $3$'s third option, where she doesn't lose $\frac\epsilon2$, and since this is favourable to Player $2$ (who gets $(1-x_1)/2$ instead of $(1-x_1/3)$), Player $2$ can play exactly at the equilibrium point and doesn't have to add a $\delta$ of his own to induce Player $3$ to choose her third option.
For larger $x_1$, it will become profitable for $x_3$ to switch to her first option. The point of indifference for this switch is $(1-x_1)/3=x_1$, or $x_1=\frac14$. For $x_1\gt\frac14$, Player $2$ plays as closely as he can to Player $1$ while still forcing Player $3$ to use her first option. The point of indifference for this is $x_2=1-x_1$, but Player $2$ has to play at $x_2=1-x_1+\delta$ to make sure that Player $3$ uses her first and not her second option. The payoffs are then $((1-2x_1)/2+\frac\delta2+\frac\epsilon2,\frac12-\frac\delta2,x_1-\frac\epsilon2)$.
We've found that by picking a number above $x_1$, Player $2$ gets $\frac12-\frac\delta2$ if $x_1\gt\frac14$ and $(1-x_1)/2\ge\frac38$ if $x_1\le\frac14$. So it will never pay for Player $2$ to play below $x_1$, and we've thus exhausted all cases.
To summarize: If Player $1$ plays at $x_1\le\frac14$, then Player $2$ plays at $x_2=(x_1+2)/3$, Player $3$ uses her third option, and the payoffs are $((5x_1+1)/6,(1-x_1)/2,(1-x_1)/3)$. If Player 1 plays at $x_1\gt\frac14$, then Player $2$ plays at $x_2=1-x_1+\delta$, Player $3$ uses her first option, and the payoffs are $((1-2x_1)/2+\frac\delta2+\frac\epsilon2,\frac12-\frac\delta2,x_1-\frac\epsilon2)$.
In the first case, the payoff for Player $1$ increases with $x_1$, and in the second case it decreases with $x_1$, so the maximum is around $\frac14$. In the first case, this yields a payoff of $\frac38$ for Player $1$, whereas in the second case the payoff is not more than $\frac14$. Thus Player $1$ picks $\frac14$, Player $2$ picks $\frac34$, and Player $3$ uses her third option, with expected payoffs $(\frac38,\frac38,\frac14)$.