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Let $\mathbb R^n$ with Euclidean distance be a Euclidean space.

Let $O=(0,0,...,0)$.

Let any line segment $AB$ from $A=(a_1,a_2,...,a_n)$ to $b=(b_1,b_2,...,b_n)$ be defined as ${\{tA+(1-t)B:0\le t \le 1}\}$.

How to prove that for any two segments $OX$ and $OY$, where $Y=\lambda X$ for some $\lambda \in \mathbb R$, either $OX \in OY$ or $OY \in OX$?

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    That's not true if you pick $Y=-X$. – 2017-02-28

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As someone mentioned in a comment, we need the condition that $\lambda \geq 0$. Also, I believe that the correct statement to prove should be that $OX \subseteq OY$ or $OY \subseteq OX$.

Given that $O = (0, 0, ..., 0)$, we can write $OX = \{ tX \; : \; 0\leq t \leq 1\}$, and similarly we can write $OY = \{ tY\; : \; 0\leq t \leq 1\}$. Since $Y = \lambda X$ for some non-negative $\lambda$, then we have $\{OY = \{ t(\lambda X) \; : \; 0\leq t \leq 1\} = \{ (t\lambda) X \; : \; 0\leq t \leq 1\} = \{ tX \; : \; 0\leq t \leq \lambda\}$, where in the last equality we made the observation that \begin{equation} \{t : 0\leq t \leq \lambda \} = \{t\lambda : 0 \leq t \leq 1\}.\end{equation}

So, we now see that $OY = \{ tX\; : \; 0\leq t \leq \lambda\}$ and $OX = \{ tX\; : \; 0\leq t \leq 1\}$. From here we conclude that $OY \subset OX$ if $0 \leq \lambda < 1$, and $OX \subset OY $ if $\lambda > 1$, and that $OX = OY$ if $\lambda = 1$.

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    How to prove that ${\{t:0 \le t \le \lambda}\}={\{t \lambda: 0 \le t \le 1}\}$? – 2017-02-28
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    @asdasdfsss Think about the numbers in each of those sets. As $t$ ranges from 0 to 1, where would $t\lambda$ range? – 2017-02-28
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    Intuitively, from $0$ to $\lambda$. – 2017-02-28
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    @asdasdfsss You're right. Which is the same of numbers as $\{t : 0\leq t \leq \lambda\}$, right? Therefore the two sets are equal. – 2017-02-28