Another equivalent way to define saturated sets: $f:X \to Y$ and $A \subseteq X$ is saturated wrt $f$ iff $f^{-1}[f[A]]=A$ or $\forall x \in X: f(x) \in f[A] \implies x \in A$.
We consider the map $\pi'_1: A \to \mathbb{R}$, the restricted projection (which is of course still continuous).
The only point that maps to $\{0\} = \pi'_1[\{(0,0)\}]$ is $(0,0)$ (because the domain is $A$!) and that already lies in $\{(0,0)\}$. So $\{(0,0)\}$ is saturated wrt $\pi'_1$. If you think about it, $\pi'_1$ is in fact 1-1 so all subsets are saturated.
Your proof is correct. However, you can do it simpler.
First observe that the restrictions $f_1 : [0,1) \to [0,2]$ and $f_2 : (2,3] \to [0,2]$ of $f$ are open maps, i.e. send open sets to open sets.
Let $U \subset [0,2]$ be a set such that $V = f^{-1}(U)$ is open. We can write $V = V_1 \cup V_2$ with open $V_1 = V \cap [0,1] \subset [0,1], V_2 = V \cap [2,3] \subset [2,3]$. Let $V'_1 = V_1 \cap [0,1)$ which is open in $[0,1)$ and $V'_2 = V_2 \cap (2,3]$ which is open in $(2,3]$. Then $f(V'_1 \cup V'_2) = f_1(V'_1) \cup f_2(V'_2)$ is open in $[0,2]$.
Case 1. $1 \notin U$. Then $1 \notin V_1, 2 \notin V_2$. Hence $V'_1 = V_1$ and $V'_2 = V_2$. Thus $V = V'_1 \cup V'_2$ and $U = f(V'_1 \cup V'_2)$ is open.
Case 2. $1 \in U$. Then $1 \in V_1, 2 \in V_2$. There is $\epsilon > 0$ such that $(1-\epsilon,1] \subset V_1, [2,2+\epsilon) \subset V_2$. Hence $V = V'_1 \cup V'_2 \cup(1-\epsilon,1] \cup [2,2+\epsilon)$ and $U = f(V) = f(V'_1 \cup V'_2) \cup f((1-\epsilon,1]) \cup f([2,2+\epsilon)) = f(V'_1 \cup V'_2) \cup (1-\epsilon,1+\epsilon)$ is open.
Edit:
Even simpler: The restrictions $f_1 : [0,1] \to [0,2]$ and $f_2 : [2,3] \to [0,2]$ of $f$ are closed maps. Let $A \subset [0,2]$ be a set such that $B = f^{-1}(A)$ is closed. We can write $B = B_1 \cup B_2$ with closed $B_1 = B \cap [0,1] \subset [0,1], B_2 = B \cap [2,3] \subset [2,3]$. Then $A = f(B) = f(B_1 \cup B_2) = f(B_1) \cup f(B_2) = f_1(B_1) \cup f_2(B_2)$ is closed.
Best Answer
An open map is a necessarily a quotient map, but the converse is not true, so your argument won't work. Use the definition: $f$ is a quotient map if it satisfies the condition: $f^{-1}(A)$ is open in $X$ if and only if $A$ is open in $Y.$ Now, note that $[0,1]$ is not open in $Y$ but $f^{-1}([0,1])$ is open in $X$, so $f$ is not a quotient map.