given two continuous functions, can I prove the inequality $||fg|| \le ||f|| ||g||$ using triangle inequality?

Question

Let $f,g : [0,1] \to $ R be continuous functions.

show that $||fg|| \le ||f||||g||$.

So what I know and found out :

The absolute value is a particular instance of the norm. So, $||x||= |x|$

Suppose I take -

$f(x)=5$ and $g(x)=2$ be continuous for all x $\in[0,1]$

Then

$|f(x)|=|5| $ and $|g(x)|=|2|$

$|fg|=|5\times2|=|10|=10$

$|f||g|=|5||2|=5\times2=10$

hence, $|fg|=|f||g|$, and from $||x||= |x|$,

we get : $||fg||=||f||||g||$

but how do I proof ? $||fg||<||f||||g||$. Am I to use triangle inequality ?