Below I specify a definition of $f_*$, and would like to directly check for its continuity. Please give your opinion on whether the proof is valid.
Let $c$ be a constant map from $X$ to $Y$ that maps $x$ to $c(x) = y$. Since $f \simeq c$, there exists a mapping $F: X \times I \to Y$ such that $F(x,0) = f(x)$ and $F(x,1) = c(x) = y$.
Now use the definition of $CX = ((X \times I)/ \sim) = (X \times [0,1)) \cup \{p\}$, where $(x_1,1) \sim(x_2,1) \ \forall x_1,x_2 \in X$, and $p$ is a point. Then define $f_*$ as: $$ \begin{cases} f_*(x,t) = F(x,t) & x \in X, 0 \leq t < 1 \\ f_*(\ p \ ) = F(x,1) = y \end{cases}$$
Let $U \subset Y$ be open. Since $F$ is continuous, $f_*^{-1}(U) = F^{-1}(U)$ is open in $X \times I$. Is it a valid assertion that $f_*^{-1}(U)$ is hence open in the quotient space $CX$?
First of all $f_* ^{-1}(U) = F^{-1}(U)$ is not true since these maps don't even have the same domain.
You have the projection map $\pi : X \times I \rightarrow CX$ and it's easy to check that $f_* \circ \pi = F$, thus we must have $\pi^{-1}(f_*^{-1}(U)) = F^{-1}(U)$.
We also know that a subset of $CX$ is open if and only if its preimage under $\pi$ is open. Thus since $F^{-1}(U)$ is open $f_*^{-1}(U)$ must also be open by definition of quotient topology, so $f_*$ is continuous.