Here's the question:
Let $f: [a,b] \to [f(a),f(b)]$ be monotonically increasing and continuous. Prove that $f$ is bijective.
Proof Attempt:
Let $f: [a,b] \to [f(a),f(b)]$ be monotonically increasing and continuous. Since it is monotonically increasing, it is injective. So, all we have to do is to prove surjectivity.
Let $K \in (f(a),f(b))$. We define the following sets:
$$E = \{x \in [a,b]: f(x) < K\}$$
Since $E$ is nonempty and bounded above, it follows that it has a least upper bound. We denote this by $c = \sup(E)$. Now, I claim that:
$$\lim_{x \to c} f(x) = K$$
This, in fact, follows from the monotonicity of $f$. If we want $|f(x)-K| < \epsilon$, then we can always choose a $\delta_1 > 0$ such that $0 < c-x < \delta_1 \implies |f(x)-K| < \epsilon$. Similarly, we can always choose a $\delta_2 > 0$ such that $0 < x-c < \delta_2 \implies |f(x)-K| < \epsilon$.
In other words, the left & right hand limits are $K$ so the limit above is $K$. Since $f$ is continuous, we conclude that $\lim_{x \to c} f(x) = K = f(c)$. However, this just shows that $f$ is surjective. Hence, $f$ is injective and surjective so it must be bijective.
Does the proof above work? If it doesn't, then why? How can I fix it? The specific thing that I'm a little unsure of is if my second last paragraph is justified or not. Like, I'm pretty sure the argument works for the left-hand limit but I'm pretty ehh about the right-hand limit.
Hint : Do you understand intuitively (I suggest a drawing) why you can get arbitrarily close to $K$ on both sides of $c$ ?