Your approach is correct and it seems to me that you have came up with an actual function between the components. Indeed, the map $C_{\alpha}\to f(C_{\alpha})$ is a bijection from the set of components of $X$ to the set of components of $Y$.
I'd like to emphasise one thing: you're constructing a map from the set of components of $X$ to the set of components of $Y$. So, $C_{\alpha}$ is an element of the former set and $f(C_{\alpha})$ is an element of the latter set. The fact that $f$ induces a bijection from $C_{\alpha}$ to $f(C_{\alpha})$ doesn't imply that $f$ induces a bijection from the set of components of $X$ to the set of components of $Y$. All it means is that it induces a map from the set of components of $X$ to the set of components of $Y$ (once you've also shown that $f(C_{\alpha})$ is a component of $Y$; see (1) below). In order to show that this induced map from the set of components of $X$ to the set of components of $Y$ is a bijection, see (2) below.
In theory, you need to prove two things (they might be obvious to you which is why you didn't explicitly note them in your proof):
(1) $f(C_{\alpha})$ is a component of $Y$
Comment: You've shown $f(C_{\alpha})$ is a connected subset of $Y$ but you also need to show that $f(C_{\alpha})$ is a maximal connected subset of $Y$. You need to use the continuity of $f^{-1}:Y\to X$ for this (something you haven't explicitly used in your proof).
(2) The map $C_{\alpha}\to f(C_{\alpha})$ is a bijection
Comment: You can show that it's a bijection by explicitly constructing a set-theoretic inverse; that is, a map from the set of components of $Y$ to the set of components of $X$. Can you do this?
Also, here's an exercise to perhaps give you an alternative perspective on this problem:
Exercise 1: Let $f:X\to Y$ be a continuous function (so, not necessarily a homeomorphism). Let $C_{X}$ and $C_{Y}$ be the sets of components of $X$ and $Y$, respectively. Prove that there is an induced map $f_{*}:C_{X}\to C_{Y}$. In the language of category theory, "the set of connected components" is a functor from the category of topological spaces to the category of sets.
The problem you're thinking about can be easily solved from the perspective of category theory as it's a general property of functors that isomorphisms are mapped to isomorphisms. (In the category of topological spaces, an "isomorphism" is just a homeomorphism; in the category of sets, an "isomorphism" is just a bijection.)
Also, the subject of homology theory or homotopy theory in algebraic topology generalises this functor; the path components of a space constitute a basis for what is known as the zeroth homology group of the space. There are higher homology groups which describe "higher dimensional holes".)
I hope this helps!
To show that $f$ is a homeomorphism, we can use the following. If $f:X\rightarrow Y$ is a continuous bijection where $X$ is compact and $Y$ is Hausdorff, then $f$ is a homeomorphism.
Then as @blamethelag suggested, you can prove continuity of $f$ by using results of product functions. (Here, I'll write $f(x)=(f_1(x),f_2(x))$.)
Let $U\times V$ be an open subset of $\mathbb{R}^2$. Then for every $x\in f^{-1}(U\times V)$, we have that $x\in f_1^{-1}(U)\cap f_2^{-1}(V)\subseteq f^{-1}(U\times V)$. Since $f_1$ and $f_2$ are continuous, $f_1^{-1}(U)\cap f_2^{-1}(V)$ is open in the domain of $f$, and this shows $f$ is continuous.
Best Answer
Surjectivity Hints:
Let $y \in \lbrack f(a),f(b) \rbrack$. Let $x_1 = \sup \{x \in \lbrack a,b \rbrack : f(x) < y\}$ and $x_2 = \inf\{x \in \lbrack a,b\rbrack : f(x) > y\}$. What's true about $x_1$ and $x_2$?
Using the answer you found to my first question and the continuity of $f$, can you prove that $f(x_1) = y$? (The easiest way to do this does not use IVT.)
This will prove that $\lbrack f(a),f(b) \rbrack \subset f(\lbrack a,b\rbrack)$. What about the opposite containment? (It's easier.)
Continuity Hints: