Nonstandardly, we can characterize a uniformly convergent sequence of functions from $E \subseteq \mathbb{R}$ by $f_n \to f$ uniformly iff $\forall x \in {}^* E . \forall n \in {}^* \mathbb{N}_\infty . f_n(x) \simeq f(x)$
Now, assuming $f_n \to f$ and $g_n \to g$ uniformly, I'm fairly sure the following proof that $f_n + g_n \to f + g$ uniformly goes through:
let $x \in {}^* E$, $n \in {}^* \mathbb{N}_\infty$. Then
$(f_n + g_n)(x) = f_n(x) + g_n(x) \simeq f(x) + g(x) = (f+g)(x)$
(where the $\simeq$ step is because of the uniform convergence of $f_n$ and $g_n$)
Now, to show that $f_n g_n \to f g$ we also need to know that $f_n$ and $g_n$ are uniformly bounded, but it feels like a similar proof, namely:
$(f_n g_n)(x) = f_n(x) g_n(x) \simeq f(x) g(x) = (fg)(x)$
does not use boundedness, but still feels correct.
There is a well known counterexample, $f_n(x) = g_n(x) = x + \frac{1}{n}$, and here $(f_n g_n)(x) = x^2 + \frac{2x}{n} + \frac{1}{n^2}$ and indeed, for $x = n \in {}^* \mathbb{N}_\infty$ we see $(f_n g_n)(x) = x^2+2 \neq x^2 = (fg)(x)$. I don't even see how this counterexample should break the proof, though...
Thanks in advance for any help!
The error is that $f_n(x)\simeq f(x)$ and $g_n(x)\simeq g(x)$ does not imply $f_n(x)g_n(x)\simeq f(x)g(x)$ in general. For instance, if $f_n(x)=\epsilon$ is a nonzero infinitesimal and $g_n(x)=1/\epsilon$, we could have $f(x)=0$ and $g(x)=1/\epsilon$, and then $f_n(x)g_n(x)=1$ but $f(x)g(x)=0$.
If you know that $f$ and $g$ are uniformly bounded, that implies $f_n(x),g_n(x),f(x),$ and $g(x)$ are all limited, which makes the deduction $f_n(x)g_n(x)\simeq f(x)g(x)$ valid. Explicitly, $f(x)g(x)-f_n(x)g_n(x)=f(x)(g(x)-g_n(x))+(f(x)-f_n(x))g_n(x)$ is infinitesimal because $g(x)-g_n(x)$ and $f(x)-f_n(x)$ are infinitesimal and $f(x)$ and $g_n(x)$ are limited.