[proofplan] We first use the Picard-Lindelöf local existence and uniqueness theorem to obtain local solutions and local uniqueness from any initial point. Then we compare any two solutions of the same initial value problem by showing that their equality set on the overlap is both relatively closed and relatively open. This compatibility allows us to define a single curve on the union of all solution intervals, and the union curve is again a solution because every point of its domain is locally represented by one of the original solutions. The construction immediately gives maximality and the stated containment property for every other solution. [/proofplan] [step:Invoke endpoint-relative local existence and local uniqueness at arbitrary solution points] We use the following standard local theorem as a prerequisite: the endpoint-relative Picard-Lindelöf local existence and uniqueness theorem. Since $F:I\times U\to\mathbb{R}^n$ is continuous and locally Lipschitz in the state variable, locally uniformly in time, for every $(\tau,\xi)\in I\times U$ there exist a number $\varepsilon_{\tau,\xi}>0$, an interval \begin{align*} L_{\tau,\xi}:=I\cap(\tau-\varepsilon_{\tau,\xi},\tau+\varepsilon_{\tau,\xi}), \end{align*} and a solution \begin{align*} u_{\tau,\xi}:L_{\tau,\xi}\to U \end{align*} of \begin{align*} \dot{u}_{\tau,\xi}(t)=F(t,u_{\tau,\xi}(t)), \qquad u_{\tau,\xi}(\tau)=\xi, \end{align*} where derivatives at endpoints of $L_{\tau,\xi}$ belonging to $L_{\tau,\xi}$ are interpreted one-sidedly. Moreover, after possibly decreasing $\varepsilon_{\tau,\xi}$, this solution is unique among all solutions with initial value $\xi$ at time $\tau$ on their common endpoint-relative subintervals of $I$. This applies at $(t_0,x_0)$, so at least one solution of the given initial value problem exists on some interval contained in $I$. [guided] The construction needs a supply of local solution pieces and a way to know that two pieces matching at one time cannot immediately separate. This is exactly the content of the Picard-Lindelöf local existence and uniqueness theorem, used here as a standard prerequisite not yet linked in the wiki. The hypotheses required by Picard-Lindelöf are the following: the vector field must be continuous in the variables $(t,x)$ and locally Lipschitz in the state variable $x$, with the Lipschitz control local and uniform over nearby times. These are precisely the hypotheses in the theorem statement. Therefore, for every point $(\tau,\xi)\in I\times U$, there is an interval $L_{\tau,\xi}\subset I$ containing $\tau$ and a map \begin{align*} u_{\tau,\xi}:L_{\tau,\xi}\to U \end{align*} such that \begin{align*} \dot{u}_{\tau,\xi}(t)=F(t,u_{\tau,\xi}(t)), \qquad u_{\tau,\xi}(\tau)=\xi. \end{align*} At endpoints of $L_{\tau,\xi}$ that belong to $L_{\tau,\xi}$, the derivative condition is interpreted one-sidedly; at interior points, it is the ordinary derivative condition. The uniqueness part is local: if another solution has the same value $\xi$ at the same time $\tau$, then the two solutions agree on some common interval around $\tau$, relative to the ambient interval $I$. This local uniqueness is the mechanism that later makes gluing well-defined. Applying the result to $(t_0,x_0)$ gives at least one local solution of the initial value problem. [/guided] [/step] [step:Show that any two solutions with the same initial data agree on their overlap] Let \begin{align*} y:K\to U \end{align*} and \begin{align*} z:L\to U \end{align*} be two solutions of the same initial value problem, where $K,L\subset I$ are intervals containing $t_0$ and $y(t_0)=z(t_0)=x_0$. Define the overlap interval \begin{align*} M:=K\cap L. \end{align*} Then $M$ is an interval containing $t_0$. Define the equality set \begin{align*} E:=\{t\in M:y(t)=z(t)\}. \end{align*} The set $E$ is nonempty because $t_0\in E$. Since $y$ and $z$ are continuous, the map \begin{align*} w:M\to\mathbb{R}^n, \qquad w(t):=y(t)-z(t), \end{align*} is continuous, and hence $E=w^{-1}(\{0\})$ is relatively closed in $M$. We next prove that $E$ is relatively open in $M$. Let $\tau\in E$, and put $\xi:=y(\tau)=z(\tau)$. By local uniqueness from the preceding step applied at $(\tau,\xi)$, there exists an interval $N\subset M$ containing $\tau$, relatively open in $M$, such that $y(t)=z(t)$ for every $t\in N$. Thus $N\subset E$, so $E$ is relatively open in $M$. Because $M$ is an interval, it is connected. The nonempty subset $E\subset M$ is both relatively open and relatively closed, so $E=M$. Hence $y(t)=z(t)$ for every $t\in K\cap L$. [/step] [step:Define the maximal interval as the union of all solution domains] Let $\mathcal{S}$ denote the set of all pairs $(K,y)$ such that $K\subset I$ is an interval containing $t_0$ and \begin{align*} y:K\to U \end{align*} is a solution of \begin{align*} \dot{y}(t)=F(t,y(t)), \qquad y(t_0)=x_0. \end{align*} The preceding local existence step shows that $\mathcal{S}\ne\varnothing$. Define \begin{align*} J:=\bigcup_{(K,y)\in\mathcal{S}} K. \end{align*} Since every $K$ in the union is an interval containing $t_0$, the set $J$ is an interval contained in $I$ and contains $t_0$. Define a map \begin{align*} x:J\to U \end{align*} as follows. For $t\in J$, choose a pair $(K,y)\in\mathcal{S}$ with $t\in K$, and set \begin{align*} x(t):=y(t). \end{align*} This definition is independent of the choice of $(K,y)$: if $(L,z)\in\mathcal{S}$ is another pair with $t\in L$, then $t\in K\cap L$, and the overlap result gives $y(t)=z(t)$. [/step] [step:Verify that the glued curve is a solution on the union interval] We prove that $x:J\to U$ is a solution of the initial value problem. Since $t_0\in K$ for every $(K,y)\in\mathcal{S}$ and every such $y$ satisfies $y(t_0)=x_0$, the definition of $x$ gives $x(t_0)=x_0$. Let $\tau\in J$. Choose $(K,y)\in\mathcal{S}$ with $\tau\in K$. We first show that there is a pair $(K_\tau,y_\tau)\in\mathcal{S}$ such that $\tau$ is an interior point of $K_\tau$ whenever $\tau$ is an interior point of $J$, and such that $K_\tau$ contains a one-sided neighbourhood of $\tau$ in $J$ whenever $\tau$ is an endpoint of $J$ belonging to $J$. If $\tau$ already has the required relative neighbourhood inside $K$, take $(K_\tau,y_\tau)=(K,y)$. Otherwise apply the endpoint-relative local existence theorem at $(\tau,x(\tau))$ and obtain a local solution \begin{align*} v:P\to U \end{align*} with $P\subset I$ an interval containing $\tau$ and $v(\tau)=x(\tau)$. Define $K_\tau:=K\cup P$. This is an interval containing $t_0$, because $K$ and $P$ are intervals and both contain $\tau$. Define \begin{align*} y_\tau:K_\tau\to U \end{align*} by $y_\tau(t)=y(t)$ for $t\in K$ and $y_\tau(t)=v(t)$ for $t\in P$. On $K\cap P$, the solution $y$ and the solution $v$ have the same value at $\tau$; local uniqueness at $(\tau,x(\tau))$ gives equality on a relative neighbourhood of $\tau$, and the open-closed overlap argument from the preceding step gives equality on all of $K\cap P$. Hence $y_\tau$ is well-defined. Away from the junction it is a solution because it agrees locally with either $y$ or $v$; at $\tau$, the left and right difference quotients, when both occur, converge to the same vector $F(\tau,x(\tau))$, so the derivative condition also holds there. Thus $(K_\tau,y_\tau)\in\mathcal{S}$. Now if $\tau$ is an interior point of $J$, choose an interval $N\subset K_\tau$ containing $\tau$ as an interior point. If $\tau$ is an endpoint of $J$ belonging to $J$, choose a one-sided interval $N\subset K_\tau$ inside $J$ with endpoint $\tau$. On $N$, the definition of $x$ agrees with $y_\tau$, so $x$ is differentiable at $\tau$ in the required ordinary or one-sided sense and \begin{align*} \dot{x}(\tau)=\dot{y}_\tau(\tau)=F(\tau,y_\tau(\tau))=F(\tau,x(\tau)). \end{align*} Therefore $x:J\to U$ is a solution on $J$. [guided] The subtle point is that a solution domain chosen merely because it contains $\tau$ need not contain a full neighbourhood of $\tau$ inside the final union $J$. For instance, $\tau$ could be the right endpoint of the particular interval $K$ even though other solution intervals continue to the right. We therefore build, when necessary, a new solution interval that does contain the required local neighbourhood of $\tau$. Choose $(K,y)\in\mathcal{S}$ with $\tau\in K$. If $K$ already contains the needed two-sided neighbourhood of $\tau$ when $\tau$ is interior to $J$, or the needed one-sided neighbourhood when $\tau$ is an endpoint of $J$, there is nothing to repair. Otherwise, apply the endpoint-relative Picard-Lindelöf theorem at the point $(\tau,x(\tau))$. Its hypotheses hold because $F$ is continuous and locally Lipschitz in the state variable, locally uniformly in time. We obtain an interval $P\subset I$ containing $\tau$ and a solution \begin{align*} v:P\to U \end{align*} with $v(\tau)=x(\tau)$. We glue $v$ to the old solution $y$. Define $K_\tau:=K\cup P$. Since both $K$ and $P$ are intervals containing $\tau$, their union is again an interval; since $K$ contains $t_0$, the interval $K_\tau$ contains $t_0$. Define \begin{align*} y_\tau:K_\tau\to U \end{align*} by setting $y_\tau(t)=y(t)$ for $t\in K$ and $y_\tau(t)=v(t)$ for $t\in P$. This definition is well-defined because on $K\cap P$ the two curves solve the same equation and have the same value at $\tau$: indeed $y(\tau)=x(\tau)=v(\tau)$. Local uniqueness first gives equality near $\tau$, and the same equality-set argument used in the preceding step then extends equality to the entire overlap $K\cap P$. The glued curve is a solution. At points belonging only to the interior of one piece, it agrees locally with that piece. At points in the overlap, the pieces are equal on a neighbourhood, so either representative gives the same derivative. At the gluing time $\tau$, if both one-sided difference quotients occur, the quotient from the $K$ side tends to $F(\tau,y(\tau))$ and the quotient from the $P$ side tends to $F(\tau,v(\tau))$; these vectors are equal because $y(\tau)=v(\tau)=x(\tau)$. Hence the ordinary derivative, or the required one-sided derivative at an endpoint, equals $F(\tau,x(\tau))$. Thus $(K_\tau,y_\tau)\in\mathcal{S}$. Now use this repaired local representative. If $\tau$ is an interior point of $J$, choose an interval $N\subset K_\tau$ containing $\tau$ as an interior point. If $\tau$ is an endpoint of $J$ belonging to $J$, choose a one-sided interval $N\subset K_\tau$ inside $J$ with endpoint $\tau$. On $N$, the definition of the union curve $x$ agrees with $y_\tau$, so \begin{align*} \dot{x}(\tau)=\dot{y}_\tau(\tau)=F(\tau,y_\tau(\tau))=F(\tau,x(\tau)). \end{align*} Therefore $x:J\to U$ satisfies the differential equation at every interior point of $J$ and in the required one-sided sense at endpoints of $J$ that belong to $J$. [/guided] [/step] [step:Prove maximality and uniqueness by the defining union property] Let \begin{align*} y:K\to U \end{align*} be any solution of the same initial value problem on an interval $K\subset I$ containing $t_0$. Then $(K,y)\in\mathcal{S}$ by definition, so \begin{align*} K\subset \bigcup_{(L,z)\in\mathcal{S}}L=J. \end{align*} For each $t\in K$, the definition of $x$ applied to the pair $(K,y)$ gives $x(t)=y(t)$. Thus every other solution is contained in $J$ and agrees with $x$ on its domain. This also proves maximality: if a solution \begin{align*} \widetilde{x}:\widetilde{J}\to U \end{align*} of the same initial value problem extends $x$ and has interval domain $\widetilde{J}\subset I$, then $(\widetilde{J},\widetilde{x})\in\mathcal{S}$, hence $\widetilde{J}\subset J$. Since an extension also has $J\subset\widetilde{J}$, we get $\widetilde{J}=J$ and $\widetilde{x}=x$. Therefore $x:J\to U$ is the unique maximal solution, and the stated strong uniqueness property follows. [/step]