Is $[a,b] \times \{c\}$ homeomorphic to $\mathbb{R^2}$? Is $[a,b] \times \{c\}$ homeomorphic to $\mathbb{R}^2$ ?
I believe this not to be the case. Take any "potential" homeomorphism $f : [a,b] \times \{c\} \to \mathbb{R^2}$. 
 Define $U \subset [a,c] \times \{c\}$ to be the set $([a,b] \times \{c\} \to \mathbb{R^2}) \ \not \ ((a,c) \cup (b,c) \cup (\frac{a+b}{2}, c)) $ .
U is disconnected, but $f(U) = \mathbb{R^2} \ \not \ (f(a,c) \cup f(b,c) \cup f(\frac{a+b}{2}))$ is connected, so $f$ can't be a homeomorphism.
Here is my issue: Although I believe my argument to hold it doesn't match with my intuition for why these two sets are not homeomorphic. A non- rigorous argument based on my intuition would be " $[a,b] \times \{c\}$ has "end points". However $ \mathbb{R^2}$ doesn't. They can't be homeomorphic".
Is there any way to make my intuition rigorous. Do my intuition and my argument even make sense/ hold true?
Any help is greatly appreciated!

If you remove a point (other than $(a,c)$ or $(b,c)$) from $[a,b]\times\{c\}$, it becomes disconnected. No point of $\mathbb R^2$ has that property.
You can also say that $[a,b]\times\{c\}$ is compact, whereas $\mathbb R^2$ isn't.
