If $S \times \Bbb{R}$ is homeomorphic to $T \times \Bbb{R}$ and $S$ is compact, can we conclude that $T$ is compact?
Since $S$ and $T$ are locally connected, we may with out loss of generality assume, that $S$ and $T$ are connected. Now $S \times \mathbb R$ and hence $T \times \mathbb R$ have exactly two ends.
A connected and locally connected space is compact iff it has no ends. Suppose $T$ is not compact, then it would have at least one end. Now the product $T \times \mathbb R$ would have exactly one end, which is the desired contradiction.