Motivation
Fix probability space $(\Omega, \mathcal F, \mathbb P)$.
Let's see how the definition of independence works in practice.
Let $ f, g: \Omega \to \mathbb R$ be measurable functions (random variables).
Denote $\mathcal A = \sigma(f) = \{ f^{-1}(E)| E\in \operatorname{Borel}(\mathbb R)\}$ and $\mathcal B = \sigma(g)$.
$\sigma$-algebras $\mathcal A$ and $\mathcal B$ are independent, if
$$
\forall A \in \mathcal A, B \in \mathcal B \quad \mathbb P (A\cap B) =
\mathbb P(A)\mathbb P(B).
$$
But elements of $\mathcal A$ are of the form $f^{-1}(E)$, for some Borel sets $E$. By convention, we write them $\{f \in E \}$. So our definition of independent variables is
$$
\forall E_1, E_2 \in \operatorname{Borel}(\mathbb R) \quad \mathbb P (f\in E_1, g\in E_2) =
\mathbb P(f\in E_1)\mathbb P(g\in E_2).
$$
where again we use some notation conventions (ommiting $\{$, $\}$ and replacing $\cap$ with $,$).
This shows what the definition is saying: any event that we could come up with
which regards $f$, is independent with any event regarding $g$. This is fairly strong condition, which comes in handy in many situations.
Examples
1. Independent but not identically distributed.
Very silly example could be $f(\omega) = 1$, and $g(\omega) = 0$ for all
$\omega \in \Omega$. Those functions are measurable in any probability space,
and all constant variables are independent (because $\sigma(f) = \sigma(g) = \{\Omega, \emptyset\}$). They obviously have different distributions.
Something more useful could be for example $\Omega = \{0, 1, 2, 3\}$, with
$$
f(x) = \begin{cases}
0 : \quad \text{when } x \leq 1\\
1 : \quad \text{when } x > 1
\end{cases}, \quad
g(x) = (x\!\!\!\! \mod 2) + 7
$$
where we deal with classical probability. Here again, distributions are quite similar (only shifted by $7$), but not the same.
Product space
In general, given 2 random variables defined on different probability spaces we can "produce probability space in which they are independent". Given $(\Omega_1, \mathcal F_1, \mathbb P_1), (\Omega_2, \mathcal F_2, \mathbb P_2), f:\Omega_1 \to \mathbb R$ and $g:\Omega_2 \to \mathbb R$ we define $\tilde f, \tilde g$ on product space $\Omega_1 \times \Omega_2$ as
$$
\tilde f(\omega_1, \omega_2) = f(\omega_1), \quad
\tilde g(\omega_1, \omega_2) = g(\omega_2).
$$
It is quite easy to show, that $\tilde f$ has the same distribution as $f$, similarly for $g$. On top of that $\tilde f$ and $\tilde g$ are independent. So you could take any 2 variables in some model (different distribution or not) and make model in which they are independent.
2. Identically distributed but not independent.
Take any symmetrical distributed variable $f$. Define $g = -f$. Then $f$ and $g$ are not independent, but since distribution of $f$ was symmetrical, we have $g \overset D = f$.
Down-to-earth version of it can be for example $f(x) = 2x-1$ on $\Omega = [0, 1]$ with Lebesgue measure as probability. $f$ has uniform distribution $U([-1, 1])$. Then $g(x) = 1 - 2x$ has the same uniform distribution. Knowing value of $f$ gives immediately value of $g$, so they are not independent (you could check that generated $\sigma$-algebras are the same, which is contradiction to $\sigma(f) \cap \sigma(g) = \{\Omega, \emptyset\}$ for any independent $f, g$).
Best Answer
To show that $f(X)$ and $f(Y)$ are identically distributed, we need to show that for any Borel set $B$, we have $P(f(X) \in B) = P(f(Y) \in B)$.
We have : $$ P(f(X) \in B) = P(\{w : f(X(w)) \in B\}) = P(\{w : X(w) \in f^{-1}(B)\}) $$
where $f^{-1}(B) = \{t \in \mathbb R : f(t) \in B\}$. Note that this set is also Borel if $B$ is, since $f$ is Borel measurable. Therefore, all the above probabilities are well defined.
Finally, note that as $X$ and $Y$ are identically distributed, $$ P(\{w : X(w) \in f^{-1}(B)\}) = P_X(f^{-1}(B))= P_Y(f^{-1}(B)) = P(\{w : Y(w) \in f^{-1}(B)\}) $$
and now go back using the logic of the first line to $P(f(Y) \in B)$ and conclude.
Now ask questions freely on this answer.