My book mentions this. However, I am not seeing why 1. and 2. are true. I guess part of the problem is that the lattice (homomorphic) map induced by the continuous topology map isn't even defined on the book (it isn't on Section 1.4.9 either). Can anyone tell me how is it usually defined, and maybe help in the subsequent statements too?
As discussed in the comments, the lattice map is the inverse set map, $f^{-1}(A) = \{ x \in X | f(x) \in A \}$ from $\tau_{2}$ to $\tau_{1}$. As also pointed out by @DanielRust in the comments, the first statement is false as he gives a valid counter-example.
My proof for the second statement would be:
Take two open sets in $\tau_{2}$, $U$ and $V$. Take $f^{-1}(U) = f^{-1}(V) = A \in \tau_{2}$ (as $f$ is continuous) Now, from the definition of $f^{-1}$, $f(A) = \{f(x) | x \in A \} \subset U, V$. Now, take $y \in U$. Because $f$ is surjective, $y = f(x')$ for some $x' \in X$. But this implies that $x' \in f^{-1}(U)$ and thus that $y \in f(A)$. Therefore, $U \subset f(A)$, which can be similarly shown for $V$. This, at once, shows that $U = f(A) = V$, and that $f^{-1}$ is surjective.
I am just beginning to self-learn set-based maths, so this is among my first such proofs I try to construct, I hope it's well done! :)