What are the strategies I can use to prove $f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T)$?

let $\ x \in f^{-1}(S \cap T)$ $\Leftrightarrow$ $f(x) \in S\cap T$

$\Leftrightarrow$ $f(x) \in S $ and $f(x) \in T $ $\Leftrightarrow$ $x \in f^{-1}(S)\cap f^{-1}(T) $

You didn’t specify the domain and codomain of $f$, so I’m going to assume that $f:X\to Y$.

Just follow the trail of definitions. Suppose that $x\in f^{-1}[S\cap T]$; by the definition of $f^{-1}$ this means that $f(x)\in S\cap T$. By the definition of intersection this means that $f(x)\in S$ and $f(x)\in T$. By the definition of $f^{-1}$ again these mean that $x\in f^{-1}[S]$ and $x\in f^{-1}[T]$. And finally, using the definition of intersection once more, we see that this means that $x\in f^{-1}[S]\cap f^{-1}[T]$. Since $x$ was an arbitrary element of $f^{-1}[S\cap T]$, we’ve shown that every element of $f^{-1}[S\cap T]$ is an element of $f^{-1}[S]\cap f^{-1}[T]$ and hence that $f^{-1}[S\cap T]\subseteq f^{-1}[S]\cap f^{-1}[T]$. (I call this kind of argument element-chasing.)

Added: I included a lot of unnecessary verbiage in an attempt to emphasize that I was doing nothing more than translating one statement into another using definitions. I wanted to emphasize that this is a ‘follow-your-nose’ proof requiring no really clever ideas: do the most obvious thing at each step, and it works. The argument can be boiled down to something very short:

Let $x\in f^{-1}[S\cap T]$. Then $f(x)\in S\cap T$, so $f(x)\in S$ and $f(x)\in T$. But then $x\in f^{-1}[S]$ and $x\in f^{-1}[T]$, so $x\in f^{-1}[S]\cap f^{-1}[T]$, and therefore $f^{-1}[S\cap T]\subseteq f^{-1}[S]\cap f^{-1}[T]$.

Now see if you can use the same sort of approach to prove the opposite inclusion.