Let $G^0 \subset G$ be the closed subgroup scheme found in Proposition \ref{proposition-connected-component}. The first two paragraphs serve to reduce to the case $G = G^0$. \medskip\noindent Observe that $T$ is a directed inverse limit of finite topological spaces (Limits, Lemma \ref{limits-lemma-topology-limit}), hence profinite as a topological space (Topology, Definition \ref{topology-definition-profinite}). Let $W \subset G$ be a quasi-compact open containing the image of $T \to G$. After replacing $W$ by the image of $G^0 \times W \to G \times G \to G$ we may assume that $W$ is invariant under the action of left translation by $G^0$, see Lemma \ref{lemma-group-scheme-over-field-translate-open}. Consider the composition $$ \psi = \pi \circ f : T \xrightarrow{f} W \xrightarrow{\pi} \pi_0(W) $$ The space $\pi_0(W)$ is profinite (Topology, Lemma \ref{topology-lemma-spectral-pi0} and Properties, Lemma \ref{properties-lemma-quasi-compact-quasi-separated-spectral}). Let $F_\xi \subset T$ be the fibre of $T \to \pi_0(W)$ over $\xi \in \pi_0(W)$. Assume that for all $\xi$ we can find an affine open $U_\xi \subset W$ with $F \subset U$. Since $\psi : T \to \pi_0(W)$ is proper as a map of topological spaces (Topology, Lemma \ref{topology-lemma-closed-map}), we can find a quasi-compact open $V_\xi \subset \pi_0(W)$ such that $\psi^{-1}(V_\xi) \subset f^{-1}(U_\xi)$ (easy topological argument omitted). After replacing $U_\xi$ by $U_\xi \cap \pi^{-1}(V_\xi)$, which is open and closed in $U_\xi$ hence affine, we see that $U_\xi \subset \pi^{-1}(V_\xi)$ and $U_\xi \cap T = \psi^{-1}(V_\xi)$. By Topology, Lemma \ref{topology-lemma-profinite-refine-open-covering} we can find a finite disjoint union decomposition $\pi_0(W) = \bigcup_{i = 1, \ldots, n} V_i$ by quasi-compact opens such that $V_i \subset V_{\xi_i}$ for some $i$. Then we see that $$ f(T) \subset \bigcup\nolimits_{i = 1, \ldots, n} U_{\xi_i} \cap \pi^{-1}(V_i) $$ the right hand side of which is a finite disjoint union of affines, therefore affine. \medskip\noindent Let $Z$ be a connected component of $G$ which meets $f(T)$. Then $Z$ has a $k$-rational point $z$ (because all residue fields of the scheme $T$ are isomorphic to $k$). Hence $Z = G^0 z$. By our choice of $W$, we see that $Z \subset W$. The argument in the preceding paragraph reduces us to the problem of finding an affine open neighbourhood of $f(T) \cap Z$ in $W$. After translation by a rational point we may assume that $Z = G^0$ (details omitted). Observe that the scheme theoretic inverse image $T' = f^{-1}(G^0) \subset T$ is a closed subscheme, which has the same type. After replacing $T$ by $T'$ we may assume that $f(T) \subset G^0$. Choose an affine open neighbourhood $U \subset G$ of $e \in G$, so that in particular $U \cap G^0$ is nonempty. We will show there exists a $g \in G^0(k)$ such that $f(T) \subset g^{-1}U$. This will finish the proof as $g^{-1}U \subset W$ by the left $G^0$-invariance of $W$. \medskip\noindent The arguments in the preceding two paragraphs allow us to pass to $G^0$ and reduce the problem to the following: Assume $G$ is irreducible and $U \subset G$ an affine open neighbourhood of $e$. Show that $f(T) \subset g^{-1}U$ for some $g \in G(k)$. Consider the morphism $$ U \times_k T \longrightarrow G \times_k T,\quad (t, u) \longrightarrow (uf(t)^{-1}, t) $$ which is an open immersion (because the extension of this morphism to $G \times_k T \to G \times_k T$ is an isomorphism). By our assumption on $T$ we see that we have $|U \times_k T| = |U| \times |T|$ and similarly for $G \times_k T$, see Lemma \ref{lemma-profinite-product-over-field}. Hence the image of the displayed open immersion is a finite union of boxes $\bigcup_{i = 1, \ldots, n} U_i \times V_i$ with $V_i \subset T$ and $U_i \subset G$ quasi-compact open. This means that the possible opens $Uf(t)^{-1}$, $t \in T$ are finite in number, say $Uf(t_1)^{-1}, \ldots, Uf(t_r)^{-1}$. Since $G$ is irreducible the intersection $$ Uf(t_1)^{-1} \cap \ldots \cap Uf(t_r)^{-1} $$ is nonempty and since $G$ is Jacobson with closed points $k$-rational, we can choose a $k$-valued point $g \in G(k)$ of this intersection. Then we see that $g \in Uf(t)^{-1}$ for all $t \in T$ which means that $f(t) \in g^{-1}U$ as desired.