Answer
We show that
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} F\left( x \right) = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} f\left( x \right)g\left( y \right) = f\left( a \right)g\left( b \right)$
Hence by definition, $F\left( {x,y} \right) = f\left( x \right)g\left( y \right)$ is continuous at $\left( {a,b} \right)$.
Work Step by Step
Let $f\left( x \right)$ be continuous at $x=a$ and $g\left( y \right)$ be continuous at $y=b$. Then, by definition
$\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)$ ${\ \ }$ and ${\ \ }$ $\mathop {\lim }\limits_{y \to b} g\left( y \right) = g\left( b \right)$
If $F\left( {x,y} \right) = f\left( x \right)g\left( y \right)$, then by Product Law of limit
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} F\left( {x,y} \right) = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} f\left( x \right)g\left( y \right) = \left( {\mathop {\lim }\limits_{x \to a} f\left( x \right)} \right)\left( {\mathop {\lim }\limits_{y \to b} g\left( y \right)} \right)$
$ = f\left( a \right)g\left( b \right)$
Since
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} F\left( x \right) = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} f\left( x \right)g\left( y \right) = f\left( a \right)g\left( b \right)$,
by definition, $F\left( {x,y} \right) = f\left( x \right)g\left( y \right)$ is continuous at $\left( {a,b} \right)$.
