Prove $f:[a,b) \longrightarrow (a,b)$ cannot be bijective and continuous (verification) I would just like to make sure my proof for this is correct. If I made any wrong assumptions, please do let me know so I can correct them or even try a different approach. Thank you!
Let $f(a)=c$, by the bijectivity of $f$. Now, consider the case where we remove $c$ from $(a,b)$ and what we are left with is a disconnected set which is the union of two connected sets, namely $(a,c) \cup (c,b)$. We now impose a restriction on $f$ defined by $$g: (a,b) \longrightarrow (a,c) \cup (c,b), \ g(x) = f(x)$$This restriction is continuous (by Theorem proved in class). The image of $g$ is the union of two connected components. Since $(a,b)$ is connected, its image must also be connected, which means it must lie entirely within one of the two components. But then this would imply that the original function, $f$, is a proper subset of $(a,b)$ which contradicts the bijectivity of it. As such, it cannot be bijective and continuous.
The intervals are regarded with the subset topology induced by the standard topology of $\mathbb R$
 A: You should specify what assumptions you use, for example that $f$ is bijective and continuous. Therefore, write "Assume that $f$ is bijective and continuous" in the beginning of the proof.
It can also be good to tell the audience that you will use a "proof by contradiction". That will make it clear why you make that assumptions on $f$.
A: The proof has a clean idea, but it's a bit confusing the way you've written it - the assumptions are a bit scattered into the proof. I would start the proof as follows, where you lay out clearly what your goal is:

We will show that, if $f:[a,b)\rightarrow (a,b)$ is continuous and injective, it cannot be surjective.

This expresses, for the reader, exactly what your proof is going to do ahead of time, rather than haphazardly bringing in assumptions during it. This helps the reader since they can verify that your proof strategy actually works without needing to parse through the details. This also reflects the structure of your proof better than a straight proof by contradiction would, since your proof is very much "we prove this property of continuous injective functions then note it's not compatible with surjectivity" rather than "we prove a property of bijective functions, then realize it's a contradiction."
Then, you can proceed:

Let $f(a)=c$.

There is no need to mention anything about $f$ here - a mere evaluation of a function does not require bijectivity, as your proof suggests. Then we can bring in the first condition we promised to assume:

By injectivity of $f$, observe that the image of $(a,b)$ under $f$ must not include $c$, hence must be a subset of $(a,c)\cup (c,b)$.

The rest of your proof is clear as is, so I won't repeat or rewrite it - you restrict the domain of $f$ to $(a,b)$, use the assumed condition of continuity to show something about connectivity, and then show that this implies that $f$ isn't surjective. Mostly, the proof you have just needs a bit of clarity at the start and a couple of improvements to be more explicit about assumptions, but is otherwise good as is.
