There is a flaw somewhere in the following argument but I can't track it.
Take a reductive connected affine algebraic group $G$ : by definition, its unipotent radical $R_u(G)$ is trivial. One have the following (non trivial) result about reductive groups :
Let $T$ be a maximal torus of $G$. Denote by $(U_\alpha)_{\alpha \in \Phi}$ an enumeration of the minimal closed unipotent normalized by $T$ subgroups of $G$. Then any closed connected subgroup $H$ of $G$ containing $T$ is generated by $T$ and the $U_\alpha$ that $H$ contains.
(Actually, the theorem is far stronger than that, but this part suffices for my purpose.)
In particular, $G$ is a closed connected subgroup of $G$ containing $T$ ; as it contains all the $U_\alpha$, one must have $$ G = \langle T, U_\alpha : \alpha \in \Phi \rangle. $$ Denote $U = \langle U_\alpha : \alpha \in \Phi \rangle$ : this the subgroup of all unipotent elements of $G$. In particular, $U$ is a unipotent subgroup of $G$ and is closed (embedding $G$ into some $\mathrm{GL}_n$, this is the subgroup of $G$ satisfying the equation $(g-1)^n=0$). It is also connected as all the $U_\alpha$ are. But it is also normal in $G$ : $T$ normalizes all the $U_\alpha$, so normalizes $U$ ; and $U$ certainly normlizes itself. So at the end, $U$ is a closed connected normal unipotent subgroup of $G$, and hence must lie into $R_u(G) = \{1\}$.
This show that every reductive group is a torus… which is certainly false.
Edit. Jack Schmidt solved it in the comments. Actually $U$ is not unipotent (and is not the set of all unipotent elements of $G$ as I said) : it is just generated by unipotent elements. The set $G_u$ of all unipotent elements of $G$ is not necessarily a group, so the product of unipotent elements is not necessarily unipotent (see the example in $SL_2$ of Jack Schmidt in the comments).
The subgroup generated by two unipotent subgroups need not be itself unipotent, since the product of two unipotent elements need not be unipotent.
For example, let $G = \operatorname{SL}_2(K)$ for some field $K$, and choose $T=\left\{ \begin{bmatrix} \lambda & 0 \\ 0 &\lambda^{-1} \end{bmatrix} : \lambda \in K^\times \right\} \cong K^\times$ to be the maximal torus. Then there are only two non-identity closed unipotent subgroups of $G$ that are normalized by $T$: $$U_1 = \left\{ \begin{bmatrix} 1 & \lambda \\ 0 & 1 \end{bmatrix} : \lambda \in K \right\} \cong K_+, \qquad U_{-1} = \left\{ \begin{bmatrix} 1 & 0 \\ \lambda & 1 \end{bmatrix} : \lambda \in K \right\} \cong K_+$$
Then $$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$$ has minimal polynomial $$(x-2)(x-1)-1 = x^2-3x+1 \neq (x-1)^2,$$ so it is not unipotent, even though it is a product of unipotent elements.
In case of $\operatorname{SL}_2$ we actually get a simpler equality: $G=\langle U_1, U_{-1} \rangle$. The subgroup $T$ is redundant here. However for $\operatorname{GL}_2$ it is needed, and also I believe for $\operatorname{PGL}_2$ over many fields.