Question about the proof of the Picard–Lindelöf theorem I am currently trying to understand the proof of the following theorem, concerning the existence and uniqueness of local solutions to an initial value problem (IVP):

Let $\Omega$ be an open subset of $\mathbb{R}\times\mathbb{R}^{n}$, $(t_{0},x_{0})\in\Omega$ a point and  $f\colon\Omega\to\mathbb{R}^{n}$ a continuous function that is Locally Lipschitz-continuous in its second variable. Then there exists a $\delta>0$ and a unique $C^{1}$-function $\hat{u}\colon[t_{0}-\delta,t_{0}+\delta]\to\mathbb{R}^{n}$ that satisfies $$(t,\hat{u}(t))\in\Omega,\quad\hat{u}'(t)=f(t,\hat{u}(t)),\quad \hat{u}(t_{0})=x_{0}$$ for all $t\in[t_{0}-\delta,t_{0}+\delta]$.

Below is a sketch of the proof, where we use the Banach fixed-point theorem:
Since $f$ is locally Lipschitz-continuous in the second variable, we are able to find an open neighbourhood $V$ around $(t_{0},x_{0})$ in $\Omega$ and a constant $L\geq0$ such that $$\|f(t,x)-f(t,y)\|\leq L\cdot\|x-y\|$$ for all $x,y\in V$. Now choose a bounded open neighbourhood $U$ of $(t_{0},x_{0})$ in $V$ such that $\overline{U}\subset V$. The boundedness of $U$ ensures that $\overline{U}$ is compact. Hence, since $f$ is continuous, the map $$\overline{U}\to\mathbb{R},\quad(t,x)\mapsto\|f(t,x)\|$$ attains a maximum $m\geq0$. Now choose $\delta>0$ and $\varepsilon>0$ such that $$[t_{0}-\delta,t_{0}+\delta]\times\overline{B_{\varepsilon}(x_{0})}\subset\overline{U},\quad m\cdot\delta\leq\varepsilon,\quad L\cdot\delta<1.$$ The space $C^{0}([t_{0}-\delta,t_{0}+\delta],\mathbb{R}^{n})$ with the metric $$d_{\max}(u,v):=\max_{|t-t_{0}|\leq\delta}\|u(t)-v(t)\|$$ is a complete metric space. Now define $$X:=\{u\in C^{0}([t_{0}-\delta,t_{0}+\delta],\mathbb{R}^{n}) \ | \ d_{\max}(u,x_{0})\leq\varepsilon\},$$ where $x_{0}$ denotes the constant function $x\mapsto x_{0}$. Then $X$ is a closed ball in the metric space $(C^{0}([t_{0}-\delta,t_{0}+\delta],\mathbb{R}^{n}),d_{\max})$. Since $X$ is closed, it is also a complete metric space (with the resticted metric). Now define a map $$\Phi\colon X\to X,\quad\Phi(u)(t):=x_{0}+\int_{t_{0}}^{t}f(\tau,u(\tau)) \ d\tau.$$ One can verify that the condition $m\cdot\delta\leq\varepsilon$ ensures that $\Phi(X)\subset X$. One can also show that $$d_{\max}(\Phi(u),\Phi(v))\leq L\cdot\delta\cdot d_{\max}(u,v)$$ So the condition $L\cdot\delta<1$ shows that $\Phi$ is a contraction. Hence, by the Banach fixed-point theorem, there exists a unique $\hat{u}\in X$ such that $\Phi(\hat{u})=\hat{u}$. Differentiating both sides with respect to $t$ yields the desired result.
MY QUESTION:  The (sketch of the) proof above shows that there exists a unique local solution $\hat{u}$ for the IVP in $X$. But the theorem states that there exists a unique solution in the space $C^{1}([t_{0}-\delta,t_{0}+\delta],\mathbb{R}^{n})$. So why is it sufficient to prove existence and uniqueness in $X$? More precisely: how do we know that the IVP has no other local solution $\hat{v}$ in $C^{1}([t_{0}-\delta,t_{0}+\delta],\mathbb{R}^{n})\setminus X$, i.e. a solution with $d_{\max}(\hat{v},x_{0})>\varepsilon$.
MY ATTEMPT: I think that it suffices to show any $u\in C^{1}([t_{0}-\delta,t_{0}+\delta],\mathbb{R}^{n})$ that solves the IVP must lie in $X$. Because such a $u$ is continuous in $t_{0}$ and satisfies $u(t_{0})=x_{0}$, there exists a $r>0$ (possibly smaller than $\delta$) such that $\|u(t)-x_{0}\|\leq\varepsilon$ whenever $|t-t_{0}|\leq r$. However, this does not prove that $\|u(t)-x_{0}\|\leq\varepsilon$ on all of $[t_{0}-\delta,t_{0}+\delta]$, i.e. that $u\in X$.
 A: So I think I found an answer to my question:
Let $\hat{v}\in C^{1}([t_{0}-\delta,t_{0}+\delta],\mathbb{R}^{n})$ be a solution of the IVP, i.e. it satisfies $$(t,\hat{v}(t))\in\Omega,\quad\hat{v}'(t)=f(t,\hat{v}(t)),\quad \hat{v}(t_{0})=x_{0}$$ for $|t-t_{0}|\leq\delta$. We prove that $\hat{v}\in X$, i.e. that $d_{\max}(\hat{v},x_{0})\leq\varepsilon$. Now define the set $$A:=\{a\in[0,\delta] \ | \ t\in[t_{0}-a,t_{0}+a]\implies\|\hat{v}(t)-x_{0}\|\leq\varepsilon\}.$$ Since $0\in A$, $A$ is non-empty. Also, $A$ is bounded from above by $\delta$, so $A$ has a supremum $s\leq\delta$. Since $\hat{v}$ is continuous in $t_{0}$ and $\hat{v}(t_{0})=x_{0}$, it follows that $s>0$. Also, by definition of $s$, we have $\|\hat{v}(t)-x_{0}\|\leq\varepsilon$ for all $t_{0}-s<t<t_{0}+s$. By continuity of $\hat{v}$ we also have $\|\hat{v}(t_{0}\pm s)-x_{0}\|\leq\varepsilon$. So, for $|t-t_{0}|\leq s$ we see that $$(t,\hat{v}(t))\in[t_{0}-\delta,t_{0}+\delta]\times\overline{B_{\varepsilon}(x_{0})})\subset\overline{U}$$ and hence $\|f(t,\hat{v}(t))\|\leq m$. Assume that $s<\delta$ and observe that $$\|\hat{v}(t_{0}\pm s)-x_{0}\|=\left\|\int_{t_{0}}^{t_{0}\pm s}f(\tau,\hat{v}(\tau)) \ d\tau\right\|\leq\left|\int_{t_{0}}^{t_{0}\pm s}\|f(\tau,\hat{v}(\tau))\| \ d\tau\right|\leq m\cdot s<m\cdot\delta\leq\varepsilon.$$ The continuity of $\hat{v}$ implies that $\|\hat{v}(t)-x_{0}\|<\varepsilon$ for $t$ in a small open interval around $t_{0}\pm s$. However, this contradicts the definition of $s$ and hence we must have $s=\delta$. This proves that $\hat{v}\in X$ and we conclude that $\hat{v}=\hat{u}$.
