You want to show $$ \| f - f_n \|_2 \to 0$$
Equivalently, you can show $$ \int (f-f_n)^2 d \mu = \| f - f_n \|_2^2 \to 0$$
We have $$ \| f - f_n \|_2^2 = \int (f-f_n)^2 d \mu = \int f^2 d \mu - 2 \int f f_n d \mu + \int f_n^2 d \mu$$
Since we have $\|f_n\|_2 \to \|f\|_2$, we have $\int f_n^2 d \mu \to \int f^2 d \mu$ and since we have $\int f_n g d \mu \to \int fg d \mu$ we have $\int f f_n d \mu \to \int f^2 d \mu$ so that
$$ \int f^2 d \mu - 2 \int f f_n d \mu + \int f_n^2 d \mu \to \int f^2 d \mu - 2 \int f^2 d \mu + \int f^2 d \mu = 0$$
that is, $$\| f - f_n \|_2^2 \to 0$$
and hence of course also
$$\| f - f_n \|_2 \to 0$$