Line segment in the unit sphere I want to prove the following statement

Let $X$ be a normed linear space, with linearly independent vectors $x,y$, such that $\|x\|=1=\|y\|$, with $\|x\|+\|y\|=\|x+y\|$, then there is a line segment in the unit sphere of $X$.

My attempt:
1. The unit sphere is the set:$$S=\{a\in X:\quad\|a\|=1\}$$
So $x,y\in S$

2. $\|x+y\|=2$, so the sum of $x$ and $y$ gives a point outside of the unit sphere, at distance two from the origin.

3. Knowing they are linearly independent, we know that $ax+by=0\iff a=b=0$

I have no idea what is meant by a line segment though, nor do I see why it is in the unit sphere. What are they wanting?
 A: Let me argue that the whole segment between $x$ and $y$ is in $S$.
For $t\in[0,1]$ set $x_t:=x+t(y-x)$. Then $x_0=x$, $x_1=y$, $x_{1/2}$ are all in $S$ by assumption: $\|x_t\|=1$ for $t\in \{0,1/2,1\}$.
By convexity of the norm, we obtain immediately that $\|x_t\|\le 1$ for all $t\in [0,1]$.
Convexity not only tells us something about the function values on the segment between two points, but also about the function values on the line outside the segment:
$$
f(\lambda u + (1-\lambda)v)\ge \lambda f(u) + (1-\lambda)f(v) \quad \forall \lambda\not\in [0,1].
$$
Applying this to points $x_t$ with $t\in (0,1/2)$ and using $\|x_{1/2}\|=\|x_1\|=1$ yields $\|x_t\|\ge 1$ for $t\in (0,1/2)$. Similar arguments work for $t\in (1/2,1)$.
Hence $\|x_t\|=1$ for all $t\in [0,1]$, and we have found a line segment in the sphere.
A: We claim that the line segment joining $x$ and $y$ is in the unit sphere:
Set $x_t=tx_1+ (1-t)x_0, t\in [0,1]$ and $x_1=x,x_0=y$. Now since the norm is convex, we have that:
$$\|x_t\| = \|tx_1+(1-t)x_0\| \leq t\|x_1\| + (1-t)\|x_0\|$$
$$\leq t+(1-t)=1$$
So we have that the line segment is in the closed unit ball, we need to show that it isn't in the open unit ball.
Assume that $\|tx_1 + (1-t)x_0\|\lt 1$ then:
$$1=\left\|\frac12 x_1 + \frac12 x_0\right\| = \left\|\frac{1-2t}{2(1-t)}x_1 + \frac{1}{2(1-t)}(tx_1+(1-t)x_0)\right\|$$
$$\implies1 = \frac{1-2t}{2(1-t)}\|x_1\| + \frac{1}{2(1-t)} \| tx_1 + (1-t)x_0)\|$$
$$\implies1 = \frac{1-2t}{2(1-t)}+ \frac{1}{2(1-t)} \| tx_1 + (1-t)x_0)\| \lt 1$$
Contradiction, hence $\|tx_1+(1-t)x_0\| \geq 1$
Hence we have  $\|tx_1+(1-t)x_0\| \geq 1,\quad t\in[0,1]$ and $\|tx_1+(1-t)x_0\| \leq 1,\quad t\in[0,1]$
It follows that $\|tx_1+(1-t)x_0\| = 1,\quad \forall t\in [0,1]$
