Reduce the compact domain of onto continuous function and it is still onto
Solution 1:
Use the poset $$\mathcal{F}= \{A \subseteq X\mid A \text{ closed in } X \land f[A]=Y\}$$ ordered by reverse inclusion. This poset is non-empty as $X$ is in it.
So $A \le B$ for $A,B \in \mathcal{F}$ iff $B \subseteq A$.
Then if $\mathcal{C} \subseteq \mathcal{F}$ is a chain, use $C_0:=\bigcap \mathcal{C}$ as an upperbound. The "challenge" is to show that $C_0 \in \bigcap{F}$ so $f[C_0]=Y$.
The restricted map (to the maximal element $X_0$) is called irreducible BTW and need not be 1-1 or a homeomorphism.