I haven't really checked the details, but I guess an argument works as follows. First, there exist $(k_i)$ such that the product map $\prod U_{k_i}\to G$ is onto. We can also arrange so that its differential map at the identity is surjective. It follows that the image of the induced map $\prod U_{k_i}(K)\to G(K)$ contains a neighborhood of the identity in the norm topology (I call $K$ the field, I don't like the notation $k$ of algebraists). Thus it has an open image. So the subgroup $V$ generated by the $U_i(K)$ is open for the norm topology.
Now first assume $G$ is simple. Then a classical result (Prasad, Bull Soc Math France 1982, who attributes it to Tits) is that any open subgroup of $G(K)$ is either compact or contains $G^+$. Since $V$ is not compact (because any of the $U_i(K)$, if not trivial, is noncompact), it follows that $V=G^+$.
In the semisimple case, the same holds: since $V$ is an open subgroup and all its projection to $K$-simple factors are noncompact, $V=G^+$. Now since $G$ is generated by unipotents, it is semisimple modulo the unipotent radical. At this point the argument is complete for $G$ reductive but still needs a little more in general. If $H$ is the quotient by the unipotent radical $W$, it is probably true, that an open subgroup of $G(K)$ whose projection to the reductive quotient $H(K)$ contains $H(K)^+$, has to contain $N(K)^+$, where $N$ is the largest perfect normal subgroup in $G$. This should allow to conclude modulo $N$, i.e. reduce to the case when $G$ itself is unipotent, which is easy.