I was given this assignment and I was wondering if my proof makes sense? Also I'd love some feedback and tips on how to improve!
Assume $f(x_{0}) = 0$, we prove $g$ is continuous on $x_{0}$. Let $ϵ > 0$. We choose δ =1. Let $|x-x_{0}|<δ$, then $\left|g\left(x\right)-g\left(x_{0}\right)\right|<ϵ$. $\left|g\left(x\right)-g\left(x_{0}\right)\right|=\left|f\left(x\right)\cdot D\left(x\right)-f\left(x_{0}\right)\cdot D\left(x_{0}\right)\right|=\left|f\left(x\right)\cdot D\left(x\right)-0\right|=\left|f\left(x\right)\cdot D\left(x\right)\right|<ϵ$
By the density of $\Bbb R\setminus\Bbb Q$ in $\Bbb R$, there exists $b∈\Bbb R$ such that $x_{0} - 1 < b < x_{0} + 1$ and $b∈\Bbb R\setminus\Bbb Q$. We choose $x = b$, then we have $\left|f\left(b\right)\cdot D\left(b\right)\right|=\left|f\left(b\right)\cdot0\right|=0<ϵ$
By the density of $\Bbb Q$ in $\Bbb R$, there exists $c∈\Bbb R$ such that $x_{0} - 1 < c < x_{0} + 1$ and $c∈\Bbb Q$. We choose x = c, then we have $\left|f\left(c\right)\cdot D\left(c\right)\right|=\left|f\left(c\right)\cdot1\right|=\left|f\left(c\right)\right|<ϵ$
Note that f is continuous at $x_{0}$, therefore $\left|f\left(c\right)-f\left(x_{0}\right)\right|<ϵ$. $\left|f\left(c\right)\right|=\left|f\left(c\right)-f\left(x_{0}\right)+f\left(x_{0}\right)\right|\le\left|f\left(c\right)-f\left(x_{0}\right)\right|+\left|f\left(a\right)\right|=\left|f\left(c\right)-f\left(x_{0}\right)\right|+\left|0\right|=\left|f\left(c\right)-f\left(x_{0}\right)\right|<ϵ$
Therefore $g$ is continuous at $x_{0}$.
Assume $g$ is continuous at $x_{0}$. We prove $f(x_{0})=0$. Note that f is continuous on $a$. Assume toward contradiction that $f(x_{0})≠0$. Then we have $g\left(x_{0}\right)=f\left(x_{0}\right)\cdot D\left(x_{0}\right)\ ➜\ \frac{g\left(x_{0}\right)}{f\left(x_{0}\right)}=D\left(x_{0}\right)$
By the algebra of continuous functions $\frac{g\left(x_{0}\right)}{f\left(x_{0}\right)}$ is continuous, but we proved in class that $D(x)$ is not continuous on any $x∈\Bbb R$. That means we have a contradiction, since we have a continuous function on one side of the equation, and a discontinuous function on the other which can't be. Therefore $f(x_{0}) = 0$.
Assume that $f(x_0) = 0$, and let $\varepsilon>0$. Then we need to find a $\delta>0$ such that for each $x \in (x_0-\delta,x_0+\delta)$ the following holds: $|f(x)D(x)| <\varepsilon$. Observe that the continuity of $f$ at $x_0$ gives us a $\delta_0 > 0$ such that $|f(x)| = |f(x)-f(x_0)|<\varepsilon$ whenever $x \in (x_0-\delta_0,x_0+\delta_0)$, so, we claim that $\delta := \delta_0$ works. Indeed, for each $x \in (x_0-\delta,x_0+\delta)$ we have two cases: either $x \in \mathbb Q$ or $x \notin \mathbb Q$.
If $x \in \mathbb Q$ then $|f(x)D(x)| = |f(x)| < \varepsilon$ because $x \in (x_0-\delta_0,x_0+\delta_0)$.
If $x \notin \mathbb Q$ then $|f(x)D(x)| = 0 < \varepsilon$.
Thus, we showed that for any $x \in (x_0-\delta,x_0+\delta)$ the following holds: $|g(x)-g(x_0)| = |f(x)D(x)| <\varepsilon$. Hence, $g$ is continuous at $x_0$.
Now suppose that $g$ is continuous at $x_0$, and for the sake of contradiction also suppose that $f(x_0) \neq 0$. Then $|f(x_0)|>0$ and the continuity of $f$ at $x_0$ gives us a $\delta>0$ such that $|f(x)-f(x_0)|<|f(x_0)|/2$ for all $x \in (x_0-\delta,x_0+\delta)$; which implies that $|f(x)| > |f(x_0)|/2$ for all $x \in (x_0-\delta,x_0+\delta)$. Thus, $f(x) \neq 0$ for all $x \in (x_0-\delta,x_0+\delta)$, and then we can define $h : (x_0-\delta,x_0+\delta) \to \mathbb R$ by $$\forall x \in (x_0-\delta,x_0+\delta) : \quad h(x) = \frac{g(x)}{f(x)}.$$ Now, by the continuity of $f$ and $g$ at $x_0$, $h$ is also continuous at $x_0$, but $h$ is the same as the restriction of $D$ to $(x_0-\delta,x_0+\delta)$, a contradiction since $D$ is discontinuous in all $\mathbb R$.