Let y be an arbitrary element.
(a) let y $\in$ f(SUT)
f$^{-1}$y $\in$ SUT
f$^{-1}$y $\in$ S or f$^{-1}$y$\in$ T
y $\in$ f(S) or y $\in$ f(T)
y $\in$ f(S)Uf(T)......means every y which belongs to f(SUT) also belongs to f(S)Uf(T), hence f(SUT) is subset of f(S)Uf(T) -(1)
let y $\in$ f(S)Uf(T)
y $\in$ f(S) or y $\in$ f(T)
f$^{-1}$y $\in$ S or f$^{-1}$y $\in$ T
f$^{-1}$y $\in$ SUT
y $\in$ f(SUT).....hence f(S)Uf(T) is subset of f(SUT) -(2)
from (1) and (2), f(SUT)=f(S)Uf(T)
(b)let y$\in$ f(S$\cap$T)
f$^{-1}$y $\in$S$\cap$T
f$^{-1}$y $\in$S and f$^{-1}$y $\in$T
y$\in$ f(S) and y $\in$f(T)
y $\in$f(S)$\cap$f(T) .....hence f(S$\cap$T) is subset of f(S)$\cap$f(T) -(1)
let y$\in$ f(S)$\cap$f(T)
y $\in$f(S) and y $\in$f(T)
f$^{-1}$y $\in$S and f$^{-1}$y$\in$ T .....this step is only true if f is one-one.It says that if y is in both f(S) and f(T) then its preimage will be common in both S and T. If f is not one one then this is false because there can be an element x1 $\in$S and x2 $\in$T such that x1$\neq$x2 and f(x1)=f(x2)=y
So f(S)$\cap$f(T) is subset of f(SUT) only if f is one-one -(2)
if f is one-one then from (1) and (2) ,f(S)$\cap$f(T) =f(SUT)
If f is not one-one then f(S$\cap$T) is subset of f(S)$\cap$f(T) and not vice versa.