Let $\omega(r)=\sup \{|f(x) - f(y)| : x,y \in B_r(a)\}$.
Suppose that $f$ is continuous. Let $\varepsilon \gt 0$. Then there is a
$\delta \gt 0$ such that
$$
|w-a| \leq \delta \Rightarrow |f(w)-f(a)| \leq \epsilon \tag{1}
$$
Let $r\leq \delta$, and let $x,y\in B_r(a)$. Using (1) with $w=x$ and $w=y$, we see
that $|f(x)-f(a)|\leq \varepsilon$ and $|f(y)-f(a)| \leq \varepsilon$. Then
$|f(x)-f(y)| \leq |f(x)-f(a)|+|f(a)-f(y)| \leq 2\varepsilon$.
We have shown
$$
r\leq \delta, x,y\in B_r(a) \Rightarrow |f(x)-f(y)| \leq 2\varepsilon \tag{2}
$$
So
$$
r \leq \delta \Rightarrow \omega(r) \leq 2\varepsilon \tag{3}
$$
This entails
$$
{\sf osc}(f,a) \leq 2\varepsilon \tag{4}
$$
Since (4) holds for any $\varepsilon \gt 0$, we deduce ${\sf osc}(f,a) \leq 0$, and hence
${\sf osc}(f,a)=0$.
Conversely, suppose ${\sf osc}(f,a)=0$. Let $\varepsilon \gt 0$. By hypothesis,
${\sf inf}_{r\gt 0}\omega(r) =0 < \varepsilon$. So there is a $\delta > 0$ such that
$\omega(\delta) < \varepsilon$. This means that :
$$
w_1,w_2\in B_{\delta}(a) \Rightarrow |f(w_1)-f(w_2)| < \varepsilon \tag{5}
$$
Let $x\in B_\delta(a)$. Then in (5) above, we may take $w_1=x,w_2=a$. This shows :
$$
|x-a| < \delta \Rightarrow |f(x)-f(a)| < \varepsilon \tag{6}
$$
So $f$ is continuous at $a$.
A product of uniformly continuous functions is not necessarily uniformly continuous. For example, set $E=\mathbb{R}$ and choose $f(x)=g(x)=x$. Their product, $x^2$, is an example of a nonuniformlycontinuous function.
Best Answer
No, it is not correct. You did not prove that there is a $\delta>0$ such that $|x|<\delta\implies|g(x)-g(0)|<\varepsilon$.
If $r=0$, then the statement is trivial. Otherwise, given $\varepsilon>0$, take $\eta>0$ such that $|x|<\eta\implies|f(x)-f(0)|<\varepsilon$. Let $\delta=\frac\eta{|r|}$. Then\begin{align}|x|<\delta&\iff|rx|<\eta\\&\implies|f(rx)-f(0)|<\varepsilon\\&\iff|g(x)-g(0)|<\varepsilon.\end{align}