Let $f,g:D\to D$ be continuous functions for every $a\in D$.
From definition of continuous functions it means that
$\forall \varepsilon >0\;\; \exists\, \delta_f \quad \forall x\in D: |x-a|\leqslant \delta_f \implies |f(x) - f(a)| \leqslant \varepsilon $,
$\forall \varepsilon >0\;\; \exists\, \delta_g \quad \forall x\in D: |x-a|\leqslant \delta_g \implies |g(x) - g(a)| \leqslant \varepsilon $.
We have to show that $f\cdot g$ is continuous at point $a$.
Note that conclusion of both implications is equivalent to
$f(a)-\varepsilon < f(x) < f(a) + \varepsilon$,
$g(a) - \varepsilon < g(x) < g(a) +\varepsilon .$
Let $\;\delta_0 := \min\{\delta_f, \delta_g\}$. Then for $|x-a| < \delta_0$ following multiplication of conclusions is true
$(f(a) -\varepsilon )(g(a) - \varepsilon ) < f(x)g(x) < (f(a) + \varepsilon )(g(a) + \varepsilon )$
$\iff f(a)g(a) - \varepsilon (f(a)-g(a)-\varepsilon ) < f(x)g(x) < f(a)g(a) + \varepsilon (f(a)+g(a)+\varepsilon )$
If $\varepsilon $ is arbitrary close to 0, then $f(x)g(x) = f(a)g(a)$.
Is this proof correct ? I have a feeling that multiplication of conclusions may not be correct.
Thanks for reading it so far.
As stated by Priyatham in comments, my multiplication is not correct, for example multiplying
$ -2< -1$ and $-3 < -2$ gives $6 < 2$, which is false.
Your proof is not correct.
In a correct proof the $\epsilon>0$ is fixed. So you cannot say things like "If $\epsilon$ is arbitrarily close to $0$ then...".
Fix some $\epsilon>0$.
Let $\delta_{1}>0$ such that $\left|f\left(x\right)-f\left(a\right)\right|\leq1$ whenever $\left|x-a\right|\leq\delta_{1}$.
Then on base of: $$\left|f\left(x\right)\right|=\left|f\left(x\right)-f\left(a\right)+f\left(a\right)\right|\leq\left|f\left(x\right)-f\left(a\right)\right|+\left|f\left(a\right)\right|\leq1+\left|f\left(a\right)\right|$$ we find that $\left|f\left(x\right)\right|\leq\left|f\left(a\right)\right|+1$ whenever $\left|x-a\right|\leq\delta_{1}$.
Let $M:=\max\left(\left|f\left(a\right)\right|+1,\left|g\left(a\right)\right|\right)$
Then on base of
$$\left|f\left(x\right)g\left(x\right)-f\left(a\right)g\left(a\right)\right|=\left|f\left(x\right)g\left(x\right)-f\left(x\right)g\left(a\right)+f\left(x\right)g\left(a\right)-f\left(a\right)g\left(a\right)\right|\leq$$$$\left|f\left(x\right)g\left(x\right)-f\left(x\right)g\left(a\right)\right|+\left|f\left(x\right)g\left(a\right)-f\left(a\right)g\left(a\right)\right|\leq\left|f\left(x\right)\right|\left|g\left(x\right)-g\left(a\right)\right|+\left|g\left(a\right)\right|\left|f\left(x\right)-f\left(a\right)\right|$$
we find that $\left|f\left(x\right)g\left(x\right)-f\left(a\right)g\left(a\right)\right|\leq M\left|g\left(x\right)-g\left(a\right)\right|+M\left|f\left(x\right)-f\left(a\right)\right|$ whenever $\left|x-a\right|\leq\delta_{1}$.
Now find $\delta_{f},\delta_{g}>0$ such that $\left|f\left(x\right)-f\left(a\right)\right|\leq\frac{1}{2}\epsilon M^{-1}$ whenever $\left|x-a\right|\leq\delta_{f}$ and $\left|g\left(x\right)-g\left(a\right)\right|\leq\frac{1}{2}\epsilon M^{-1}$ whenever $\left|x-a\right|\leq\delta_{g}$ .
Then for $\delta=\min\left(\delta_{1},\delta_{f},\delta_{g}\right)>0$ we find that $\left|f\left(x\right)g\left(x\right)-f\left(a\right)g\left(a\right)\right|\leq\epsilon$ whenever $\left|x-a\right|\leq\delta$.