real analysis - If $f,g$ are uniformly continuous prove $f+g$ is uniformly continuous but $fg$ and $dfrac{f}{g}$ are not

Suppose $f:\mathbb{R} \supset E \rightarrow \mathbb{R}$ and $g: \mathbb{R} \supset E \rightarrow \mathbb{R}$ are uniformly continuous. Show that $f+g$ is uniformly continuous. What about $fg$ and $\dfrac{f}{g}$?

My Attempt

Firstly let's state the definition; a function is uniformly continuous if
$$\forall \varepsilon >0\ \ \exists \ \ \delta >0 \ \ \text{such that} \ \ |f(x)-f(y)|< \varepsilon \ \ \forall \ \ x,y \in \mathbb{R} \ \ \text{such that} \ \ |x-y|<\delta$$

Sum $f+g$

Now to to prove $f+g$ is uniformly continuous;
$\bullet$ Choose $\delta_1 >0$ such that $\forall$ $x,y \in \mathbb{R}$ $|x-y|<\delta_1$ $\implies$ $|f(x)-f(y)|< \dfrac{\epsilon}{2}$

$\bullet$ Choose $\delta_2 >0$ such that $\forall$ $x,y \in \mathbb{R}$ $|x-y|<\delta_2$ $\implies$ $|g(x)-g(y)|< \dfrac{\varepsilon}{2}$

$\bullet$ Now take $\delta := min\{ \delta_1, \delta_2\}$ Then we obtain for all $x,y \in \mathbb{R}$
$$
|x-y|<\delta \implies

|f(x)+g(x)-f(y)+g(y)| <
|f(x)-f(y)| + |g(x)-g(y)| <
\dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}=
\varepsilon$$

---

Product $fg$

Now for $fg$ for this to hold both $f:E \rightarrow \mathbb{R}$ and $g:E \rightarrow \mathbb{R}$ must be bounded , if not it doesn't hold.

$\bullet$ $\exists \ \ M>0 \ \ such \ that \ \ |f(x)|

$\bullet$ Choose $\delta_1 >0$ such that $\forall$ $x,y \in \mathbb{R}$ $|x-y|<\delta_1$ $\implies$ $|f(x)-f(y)|< \dfrac{\epsilon}{2M}$

$\bullet$ Choose $\delta_2 >0$ such that $\forall$ $x,y \in \mathbb{R}$ $|x-y|<\delta_2$ $\implies$ $|g(x)-g(y)|< \dfrac{\epsilon}{2M}$

$\bullet$ Now take $\delta := min\{ \delta_1, \delta_2\}$. Then, $|x-y|<\delta$ implies for all $x,y \in \mathbb{R}$, that
$$|f(x)g(x)-f(y)g(y)| \leq
|g(x)||f(x)+f(y)|+|f(y)||g(x)+g(y)| \leq
$$

$$
M|f(x)+f(y)| + M|g(x)+g(y)| <
M \dfrac{\epsilon}{2M} + M \dfrac{\epsilon}{2M} =
\epsilon$$

Are these proofs correct?
I am not sure how to approach the $\dfrac{f}{g}$ case.