Uniqueness of endpoints of half-open line segments in linear spaces. I try to solve the following exercise, which is Exercise 1.18 in Robert Megginson's An Introduction to Banach Space Theory.

Let $X$ be a linear space, and define for any $x_1, x_2 \in X$ the line segment
  $$ (x_1, x_2] := \{y \in X \;| \; y = (1-t)x_1 + tx_2, 0 < t \leq 1\}. $$
  Suppose that $(x_1, x_2] = (y_1, y_2]$. Then $x_1 = y_1$ and $x_2 = y_2$.

The case $x_1 = x_2$ is trivial, but I am not sure how to do the case $x_1 \neq x_2$. Any help is appreciated. 
EDIT: The exercise contains the following hint:

It may help to show first that $[x_1, x_2] = [y_1, y_2]$, where 
  $$[x_1, x_2] = \{y \in X \;| \; y = (1-t)x_1 + tx_2, 0 \leq t \leq 1\}.$$ 
  To this end, notice that $(x_1, x_2] = [x_1, x_2]$ are both convex and that $[x_1, x_2]$ has exactly one more point than $(x_1, x_2]$ does. In how many ways can $(x_1, x_2]$ be augmented by one point so that the resulting set is convex?

 A: Here is a solution, sorry that it is rather long, I'm sure a lot can be cut.
Let $x_1,x_2 \in [y_1,y_2]$, then since $[x_1,x_2]$ is the convex span of $x_1$ and $x_2$ you have $[x_1,x_2] \subset [y_1,y_2]$. Similarly $y_1,y_2 \in [x_1,x_2]$ implies $[y_1,y_2] \subset [x_1,x_2]$.
In the following $\alpha, \tilde \alpha, \beta, \gamma \in [0,1]$. Assume $[y_1,y_2]=[x_1,x_2]$, expand $y_1$ in $x_1,x_2$ and the resulting expression again in $y_1,y_2$.
$$y_1=\alpha x_1 + (1-\alpha)x_2=\alpha \beta y_1+\alpha (1-\beta)y_2+(1-\alpha)\gamma y_1+(1-\alpha)(1-\gamma)y_2$$
Assuming $y_1, y_2$ to not be proportional to each other (linearly independent) gives then
$$\alpha\beta+\gamma-\alpha\gamma=1 \tag{1}$$
Noting $$y_2=\tilde \alpha x_1+ (1-\tilde\alpha)x_2=\tilde\alpha \beta y_1+\tilde\alpha (1-\beta)y_2+(1-\tilde\alpha)\gamma y_1+(1-\tilde\alpha)(1-\gamma)y_2$$
Gives $$\tilde \alpha \beta + \gamma -\tilde \alpha \gamma =1 \tag{2}$$
Subtracting $(1)$ and $(2)$ gives $(\tilde\alpha+\alpha)(\beta-\gamma)=0$, and either $\tilde\alpha=\alpha=0$ or $\beta=\gamma$ from which $\gamma=1$ follows from $(1)$.
So either $y_1=x_2$ and $y_2=x_1$ or $y_1=x_1$ (and then $y_2=x_2$), so if $[x_1,x_2]=[y_1,y_2]$ you have the endpoints all agree.
Now if $(x_1,x_2]=(y_1,y_2]$ then adding the point $x_1$ on the left set gives you $[x_1,x_2]$, which remains convex. The thing that remains to be seen is that the only point you can add to $(y_1,y_2]$ so that the set remains convex is $y_1$. To that end let $\alpha\in [0,1),\beta \in (0,1]$ and let $y_3 \in X$, $y_3 \notin (y_1,y_2]$. If $\{y_3\}\cup(y_1,y_2]$ is convex, then:
$$\alpha y_3 + (1-\alpha)\beta y_1 + (1-\alpha)(1-\beta)y_2$$
must lie in $(y_1,y_2]$ (as $\alpha \neq1, \beta \neq 0$). At any rate you get $y_3 \in \rm{span}\{y_1,y_2\}$. So $y_3= a y_1 + b y_2$ and:
$$\alpha(a-\beta)y_1+\beta y_1+(1-\alpha-\beta+\alpha\beta +\alpha b)y_2$$
is in $(y_1,y_2]$ for $\alpha, \beta \in (0,1]$. If $y_1,y_2$ are linearly independent you get that (since it must be of form $ty_1+(1-t)y_2$):
$$\alpha(a-\beta)+\beta=\alpha+\beta-\alpha\beta-\alpha b\tag{3}$$
or
$\alpha(a+b)=\alpha$ and $a=1-b$. Now both sides of $(3)$ must be in $(0,1]$ for $\alpha, \beta \in (0,1]$. From continuity of the expressions extend to $[0,1]$, then $\alpha=1, \beta=0$ gives on the right: $1-b$ must be in $[0,1]$, so $0≤b≤1$ and thus $y_3=(1-b)y_1+by_2$ is in $[y_1,y_2]$.
But the only point in $[y_1,y_2]$ that is not in $(y_1,y_2]$ is $y_1$. So $x_1=y_1$. Finally $[x_1,x_2]=\{x_1\}\cup(x_1,x_2]=\{y_1\}\cup(y_1,y_2]=[y_1,y_2]$ and the previous considerations give $x_2=y_2$ also.
