Composition of quotient maps is a quotient map
My attempt:
Let $q_1:X\rightarrow Y$, $q_2:Y\rightarrow Z$ be quotient maps. I will show that $q_2\circ q_1$ is a quotient map. Since the composition of surjections is a surjection, the composition is a surjection. Let $U$ be open in $Z$. So $q_2^{-1}(U)$ is open in $Y$ and since $q_1$ is continuous, $q_1^{-1}(q_2^{-1}(U))$ is open in $X$. This implies that $(q_2\circ q_1)^{-1}(U)$ is open in $X$. Hence $Z$ admits the quotient topology, so $q_2\circ q_1$ is a quotient map.
Is my proof correct and complete? (Please answer this)
No, you’re proving the wrong direction and reproving that the composition of continuous maps is continuous. So we now know that $q_2 \circ q_1 $ is continuous and indeed onto.
To see the quotient map property, we assume that $O \subseteq Z$ is such that $(q_2 \circ q_1)^{-1}[O]$ is open. Note then again that
$$(q_2 \circ q_1)^{-1}[O]=q_1^{-1}[q_2^{-1}[O]] \text{ is open}$$
And as $q_1$ is quotient we conclude that
$$q_2^{-1}[O] \text{ is open}$$ and finally, as $q_2$ is quotient, we conclude that $O$ is open as required.