1. Setup and Assumptions:
Suppose there are $n$ parallel segments $A_1B_1, A_2B_2, \ldots, A_nB_n$ in the plane, all parallel to the x-axis. Assume that the segments are ordered such that $A_kB_k$ is above $A_{k+1}B_{k+1}$ for all $k$, and $A_k$ is the left endpoint while $B_k$ is the right endpoint of the segment.

2. **Initial Line $L_1$**:
Consider a line $L_1$ that joins $B_1$ and $B_m$ for some $m$. We choose $m$ such that all points $B_i$ (for $i = 1, 2, \ldots, n$) are either on $L_1$ or to the right of $L_1$. This is possible because if any $B_i$ lies to the left of $L_1$, we can choose $m = i$ and repeat the process until all $B_i$ are to the right of $L_1$.

3. **Position of $A_i$ Relative to $L_1$**:
Now, if there exists an $i < m$ such that $A_i$ lies to the right of $L_1$, then there would be no line passing through the segments $A_1B_1, A_iB_i, A_mB_m$, which contradicts the given condition. Therefore, all $A_i$ (for $i < m$) must lie to the left of $L_1$.

4. **Line $L_2$**:
Next, consider a line $L_2$ that joins $B_m$ and $A_k$ for some $k < m$. We choose $k$ such that all points $A_i$ (for $i < m$) lie to the left of $L_2$.

5. **Position of $B_i$ Relative to $L_2$**:
If there exists a $B_i$ (for $i > m$) that lies to the left of $L_2$, we consider a line $L_3$ that joins $B_m$ and $B_j$ for some $j > m$. We choose $j$ such that all points $B_i$ (for $i > m$) lie to the right of $L_3$. If no such $B_i$ exists, let $L_3 \equiv L_2$.

6. **Final Line $L_3$**:
Now, all $B_i$ (for $i > m$) lie to the right of $L_3$, and all $A_i$ (for $i < m$) lie to the left of $L_3$. Thus, $L_3$ intersects all the segments $A_1B_1, A_2B_2, \ldots, A_nB_n$.

Therefore, we have shown that there exists a line that intersects all the given segments.

$\blacksquare$