Proof by contradiction
Let's assume $\space \exists \space f(n) \in o(g(n)) \space \cap \omega(g(n)) $.
By definition of $f(n) = o(g(n))$, we have,
$f(n) < c \cdot g(n) \space \forall \space c \in R_{0} $
where $R_{0}$ is a set of positive real numbers.
Now, by definition of $f(n) = \omega(g(n))$, we have,
$f(n) > c \cdot g(n) \space \forall \space c \in R_{0} $
Combining above two inequalities,
$f(n) < c \cdot g(n) < f(n)$ which is a contradiction, since a value cannot be strictly greater and smaller than another value, at the same time.
Hence, the above inequality does not hold for all positive real values of constant $c$.
Hence, $o(g(n)) \space \cap \omega(g(n)) = \emptyset $.
Another way to prove by using the limits definitions
Let's assume $\space \exists \space f(n) \in o(g(n)) \space \cap \omega(g(n)) $.
Which means, $f(n) \in o(g(n))$ and $f(n) \in \omega(g(n))$.
Now, according to definition of $f(n) \in o(g(n))$,
$\lim_{n \rightarrow \infty }{\frac{g(n)}{f(n)} } = \infty$
which means, $\lim_{n \rightarrow \infty }{\frac{1}{\frac{g(n)}{f(n)} } }= 0$
which means, $\lim_{n \rightarrow \infty }{\frac{f(n)}{g(n)} }= 0$
Now, according to the definition of $f(n) \in \omega(g(n))$,
$\lim_{n \rightarrow \infty }{\frac{f(n)}{g(n)} }= \infty$
Which is a contradiction since one definition says the limit evaluates to infinity, while other says it evaluates to zero.
Hence, no such function $f(n)$ exists.
Hence, $o(g(n)) \space \cap \omega(g(n)) = \emptyset $.