This strategy obtains $x = 1.5$. (All unitless distances will be in kilometers.) Step 1: Have Alice walk forward 0.5 while Bob waits. Step 2: Have Bob walk forward to the front of the train while Alice waits. Step 3: Have Alice walk forward to the front of the train while Bob waits. (Of course, we stop early if Alice and Bob meet.) To show that this never exceeds 1.5, we take two cases. Case 1: Alice starts in front of Bob. Then step 2 will have Bob meet Alice. Alice has walked at most 0.5 and Bob has walked at most 1, so we're okay. Case 2: Bob starts in front of Alice. Let $a$ and $b$ be the respective distances of Alice and Bob from the front of the train. If $b > 0.5$, Alice will find Bob in step 1 and the total distance is less than 0.5. If $b \leq 0.5$, then $a + b \leq 1.5$ and completing all of steps 1-3 will give a total distance of $a+b$.

Let's suppose a strategy exists with $x < 1.5$. Let $\delta = 1.5 - x > 0$. We will find a contradiction. (All unitless distances will be in kilometers.) We'll choose the actual initial conditions Alice and Bob deal with as late as possible, and have them follow their strategies while delaying them finding any new information (meeting or reaching either end of the train). Suppose at any given point in time, the point closest to the back of the train that Alice has searched is $A_-$, and the point closest to the front is $A_+$. Similarly define $B_-$ and $B_+$. We haven't specified the exact locations of any of these points yet, but we (and Alice and Bob) know how far Alice is currently from $A_-, A_+$ and how far Bob is from $B_-, B_+$. While $A_-A_+ + B_-B_+ = 1 - \epsilon$ for $\epsilon > 0$, two situations are possible where Alice and Bob have learned nothing. One is that $A_+$ is $\epsilon/3$ away from the front of the train, $B_-$ is $\epsilon/3$ away from the back of the train, and $B_+$ is behind $A_-$ by $\epsilon/3$. The same is possible with Alice and Bob switched. So there is no way for them to know which way to walk to reach each other, and we'll leave the actual situation undetermined. For convenience, we'll set $c = x/4 + 1/8$. The most important things about this quantity are that $2c = 1 - \delta/2 < 1$ and $x < 3c < 3/2$. The contradiction we'll obtain is that the strategy must walk a distance of $3c$ or more. Take the first time $T$ at which $A_-A_+ + B_-B_+\geq 2c$. Without loss of generality, the time $T$ occurs while Alice is at $A_+$, meaning she was exploring new parts of the train in the forwards direction at $T$. Let Bob's current position be $B_1$, so $B_1$ is between $B_-$ and $B_+$ inclusive. At time $T$, note the total distance walked so far is at least $2c$, since $A_-A_+ + B_-B_+\geq 2c$. We claim we can set things up so that $B_1A_+ \geq c$. We'll now decide what the actual starting positions were with a couple cases. Case 1: $B_1B_+ + A_-A_+ \geq c$. We'll set the initial positions so Alice just reached the front of the train, $B_-$ was $\delta/4$ away from the back of the train, and $B_+$ was behind $A_-$ by $\delta/4$. The distance between Alice and Bob is $B_1A_+ = B_1B_+ + B_+A_- + A_-A_+$, which is at least $c$ by assumption. Case 2: $B_-B_1 \geq c$. We'll set the initial positions so that Alice just reached $B_-$ (i.e. $A_+ = B_-$), $B_+$ was $\delta/4$ away from the front of the train, and $A_-$ was $\delta/4$ away from the back of the train. The distance between Alice and Bob is $A_+B_1 = B_-B_1$, which is at least $c$ by assumption. Because $B_-B_1 + B_1B_+ + A_-A_+ = A_-A_+ + B_-B_+ \geq 2c$, at least one of the two cases above applies. In either case, we find $B_1A_+ \geq c$, which means that the total walking distance before meeting is at least \[ A_-A_+ + B_-B_+ + B_1A_+ \geq 3c. \]Since $3c > x$, the bound of the hypothetical strategy, we have arrived at a contradiction. So $x < 1.5$ is not possible.