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 $\mathbb R$*