An Open Interval and a Half-open Interval are not Homeomorphic As the topic, how can I prove it?

Prove that (a,b) and [a,b) in R are not homeomorphic metric space.

I'm pretty weak on this section (homeomorphism) and I don't have any clue to figure it out.
Please help me, thank you.
 A: Let $f: [a,b) \to (a,b)$ be a homeomorphism. Then the image of $(a,b)$ is $(a,b) \setminus \{f(a)\}$. Because $(a,b)$ is connected and $f$ is continuous, the image of $(a,b)$ is connected.
Can $(a,b) \setminus \{f(a)\}$ be connected (we know that $f(a)$ is in $(a,b)$ )? Assume it is. Now $(a,b)\setminus \{f(a)\} = (a, f(a)) \cup (f(a) ,b)$. What does this mean?
A: Without going into formalisms, if two spaces are homeomorphic then all of their topological properties are the same.
Now the space $(a,b)$ has the (topological) property that if you remove any of its points, the resulting space is disconnected. But the space $[a,b)$ does not have this property -- why?
A: Thank to all above, now I write it this way and it seems complete.

Suppose not, i.e. if there exist $f:(a,b) \to [a,b)$ a homeomorphism.
By definition, f is continuous, $f^{-1}$ is continuous, and $f$ is bijective.
$f$ is bijective then we have the inverse image of $[a,b)$ is $(a,b)$
$f^{-1}(a)$ is a single point, say c, and $a \lt c \lt b$.
Consider $f^{-1}((a,b))=(a,c) \cup (c,b)$ is disconnected.
It contradicts the fact that $f^{-1}((a,b))$ is connected, noted that $(a,b)$ connected and $f^{-1}$ is continuous.
Therefore, $(a,b)$ and $[a,b)$ are not homeomorphic.

