Weak convergence argue

Since the term $g(y)$ doesn’t show up in the limit, I will assume you mean $$ \int_{\Omega}g_n(x)\phi(x)\int_\Omega f(x,y)u_n(y)dydx\to \int_{\Omega}g(x)\phi(x)\int_\Omega f(x,y)u(y)dydx $$ (or perhaps $\int_{\Omega}g_n(y)\phi(x)\int_\Omega f(x,y)u_n(x)dydx\to \int_{\Omega}g(y)\phi(x)\int_\Omega f(x,y)u(x)dydx$)? Otherwise the result looks false at first glance. Note that the operator $Su=\int_\Omega f(x,y)u(y)dy$ is an example of a Hilbert-Schmidt operator on $L^2$. It is thus compact, and so $Su_n \to Su$ strongly in $L^2$. Thus, we have $$ \int_\Omega g_n(x)\phi(x)S u_n(x)dx = \int_{\Omega} g_n(x)\phi(x)Su(x)dx +\int_\Omega g_n(x)\phi(x) (Su_n-Su)(x)dx $$ The first term converges to zero since $\phi Su \in L^1$ by Cauchy-Schwarz, and the second term is bounded above by $$ ||g_n||_{L^\infty} ||\phi||_{L^2}||Su_n-Su||_{L^2} \to 0$$ yielding the result. Note that if you want to go the route you were trying, you need to get stronger convergence. Indeed, note that the sequence $$ Tg_n=\int_\Omega f(x,y)g_n(y)dy $$ will in fact converge in the supremum norm to $Tg$. Indeed, $f$ is continuous on a compact set, and thus uniformly continuous. Thus, given $\epsilon>0$, there exists $\delta >0$ so that $|(x,y)-(z,w)|<\delta$ implies $|f(x,y)-f(z,w)|<\epsilon$. Hence, for $|x-z|<\delta$, we have that $$ |Tg_n(x)-Tg_n(z)|\leq \int_\Omega |f(x,y)-f(z,y)||g_n(y)|dy \leq \epsilon ||g_n||_{L^\infty}|\Omega| $$ showing equicontinuity of the sequence $Tg_n$. Hence, by Arzela-Ascoli, this has a uniformly convergent subsequence, and since we have pointwise converge too, it actually converges uniformly without restricting to a subsequence. Hence, we have that $$ \int_{\Omega}u_n(x)\phi(x) Tg_n(x)dx =\int_\Omega u_n(x)\phi(x)Tg_n(x)-Tg(x)dx +\int_{\Omega}u_n(x)\phi(x)Tg(x)dx $$ which converges to zero by similar reasoning as before.