[proofplan] We prove that the complement of the closed ball is open in the norm topology. If a point $y$ lies outside $\overline{B}(x,r)$, then the positive gap $\|y-x\|-r$ gives a norm-open ball around $y$. The triangle inequality shows that every point in this smaller ball still has distance greater than $r$ from $x$, so the whole neighbourhood is contained in the complement. [/proofplan] [step:Fix an exterior point and define its distance gap] Fix $x\in V$ and $r\in [0,\infty)$. By definition of the closed ball, \begin{align*} \overline{B}(x,r)=\{w\in V:\|w-x\|\le r\}. \end{align*} Let $y\in V\setminus \overline{B}(x,r)$. Then $\|y-x\|>r$. Define the positive number \begin{align*} \varepsilon:=\|y-x\|-r. \end{align*} Since $\|y-x\|>r$, we have $\varepsilon>0$. [/step] [step:Show the $\varepsilon$-ball around the exterior point stays outside the closed ball] We claim that the norm-open ball \begin{align*} B(y,\varepsilon):=\{z\in V:\|z-y\|<\varepsilon\} \end{align*} is contained in $V\setminus \overline{B}(x,r)$. Let $z\in B(y,\varepsilon)$. By the triangle inequality applied to \begin{align*} y-x=(y-z)+(z-x), \end{align*} we obtain \begin{align*} \|y-x\|\le \|y-z\|+\|z-x\|. \end{align*} Rearranging gives \begin{align*} \|z-x\|\ge \|y-x\|-\|y-z\|. \end{align*} Since $\|y-z\|=\|z-y\|<\varepsilon=\|y-x\|-r$, it follows that \begin{align*} \|z-x\|>\|y-x\|-(\|y-x\|-r)=r. \end{align*} Thus $z\notin \overline{B}(x,r)$. Since $z\in B(y,\varepsilon)$ was arbitrary, \begin{align*} B(y,\varepsilon)\subset V\setminus \overline{B}(x,r). \end{align*} [guided] We want to prove that the complement of the closed ball is open. In the norm topology, this means that for each point $y$ outside the closed ball, we must produce a positive radius $\varepsilon$ such that every point sufficiently close to $y$ is also outside the closed ball. Because $y\notin \overline{B}(x,r)$, the defining condition for the closed ball fails. Hence \begin{align*} \|y-x\|>r. \end{align*} The strict inequality gives a positive margin between the distance from $y$ to $x$ and the radius $r$. Define \begin{align*} \varepsilon:=\|y-x\|-r. \end{align*} Then $\varepsilon>0$. Now let $z\in V$ satisfy $\|z-y\|<\varepsilon$. We must prove that $z$ is still outside the closed ball, meaning that $\|z-x\|>r$. The only estimate needed is the triangle inequality. Since \begin{align*} y-x=(y-z)+(z-x), \end{align*} the triangle inequality gives \begin{align*} \|y-x\|\le \|y-z\|+\|z-x\|. \end{align*} Subtracting $\|y-z\|$ from both sides yields \begin{align*} \|z-x\|\ge \|y-x\|-\|y-z\|. \end{align*} The hypothesis $\|z-y\|<\varepsilon$ is the same as $\|y-z\|<\varepsilon$, because a norm satisfies $\|y-z\|=\|-(z-y)\|=\|z-y\|$. Therefore \begin{align*} \|z-x\|>\|y-x\|-\varepsilon. \end{align*} Substituting the definition $\varepsilon=\|y-x\|-r$ gives \begin{align*} \|z-x\|>\|y-x\|-(\|y-x\|-r)=r. \end{align*} Thus $z\notin \overline{B}(x,r)$. Since every $z$ with $\|z-y\|<\varepsilon$ has this property, \begin{align*} B(y,\varepsilon)\subset V\setminus \overline{B}(x,r). \end{align*} [/guided] [/step] [step:Conclude that the complement is open] For every $y\in V\setminus \overline{B}(x,r)$, we have found $\varepsilon>0$ such that \begin{align*} B(y,\varepsilon)\subset V\setminus \overline{B}(x,r). \end{align*} Therefore $V\setminus \overline{B}(x,r)$ is open in the norm topology. Hence $\overline{B}(x,r)$ is closed in the norm topology on $V$. [/step]