[proofplan] We show that $B(x, r) = \{y \in K : |y - x| \leq r\}$ is open in the metric topology by exhibiting an open ball around each of its points that is contained in $B(x, r)$. The key ingredient is the [Every Point Is a Center](/theorems/???) property: once $z \in B(x, r)$, the equality $B(z, r) = B(x, r)$ holds by the ultrametric inequality, and in particular the open ball $B^\circ(z, r) = \{y \in K : |y - z| < r\} \subseteq B(z, r) = B(x, r)$. Since $B^\circ(z, r)$ is an open neighbourhood of $z$ contained in $B(x, r)$, every point of $B(x, r)$ is interior, and $B(x, r)$ is open. This phenomenon is peculiar to non-archimedean fields: in $\mathbb{R}$, closed balls are not open. [/proofplan]

[step:Show every point of $B(x, r)$ is an interior point]Let $z \in B(x, r)$, so $|z - x| \leq r$. We claim that the open ball \begin{align*} B^\circ(z, r) := \{y \in K : |y - z| < r\} \end{align*} is an open neighbourhood of $z$ contained in $B(x, r)$. For any $y \in B^\circ(z, r)$, we have $|y - z| < r$, so in particular $|y - z| \leq r$, meaning $y \in B(z, r)$. By the [Every Point Is a Center](/theorems/???) theorem — which applies because $z \in B(x, r)$ and $|\cdot|$ is a non-archimedean absolute value — we have $B(z, r) = B(x, r)$. Therefore $y \in B(x, r)$. This shows $B^\circ(z, r) \subseteq B(x, r)$, so $z$ is an interior point of $B(x, r)$. Note that the radius of the open neighbourhood $B^\circ(z, r)$ is $r$, independent of $z$. In the archimedean case, a boundary point $z$ with $|z - x| = r$ exactly has no open ball contained in $B(x, r)$. In the non-archimedean setting, there is no meaningful "boundary" — every point satisfying $|z - x| \leq r$ has a full-radius open neighbourhood inside the ball.[/step]

[guided]The goal is to show that the closed ball $B(x, r) = \{y \in K : |y - x| \leq r\}$ is an open set in the topology induced by the absolute value $|\cdot|$. A set is open if every point has an open neighbourhood contained in the set. Let $z \in B(x, r)$. We need to find $\varepsilon > 0$ such that $B^\circ(z, \varepsilon) := \{y \in K : |y - z| < \varepsilon\} \subseteq B(x, r)$. The natural choice is $\varepsilon = r$ itself. Take any $y \in B^\circ(z, r)$, so $|y - z| < r$. In particular $|y - z| \leq r$, which means $y \in B(z, r)$. By the [Every Point Is a Center](/theorems/???) theorem, since $z \in B(x, r)$ and $(K, |\cdot|)$ is non-archimedean, we have $B(z, r) = B(x, r)$. Therefore $y \in B(x, r)$. We have shown $B^\circ(z, r) \subseteq B(x, r)$, so $z$ is an interior point of $B(x, r)$. Why does this fail in archimedean fields? In $\mathbb{R}$ with the standard absolute value, the closed ball $[0, 1]$ is not open because the boundary point $1$ has no open neighbourhood contained in $[0, 1]$. The "every point is a center" property fails in archimedean fields, which is what makes non-archimedean geometry fundamentally different.[/guided]

[step:Conclude that $B(x, r)$ is open]Since every point $z \in B(x, r)$ is an interior point (with $B^\circ(z, r)$ serving as the required open neighbourhood), the set $B(x, r)$ is open in the metric topology induced by $|\cdot|$. Explicitly: the open cover $B(x, r) = \bigcup_{z \in B(x, r)} B^\circ(z, r)$ expresses $B(x, r)$ as a union of open balls, confirming it is open by the definition of the metric topology (arbitrary unions of open balls are open). Note that we chose $\varepsilon = r$ for every interior point — the same radius works everywhere. This uniformity is a direct consequence of the [Every Point Is a Center](/theorems/???) property: once we know $B(z, r) = B(x, r)$, the open ball $B^\circ(z, r) \subset B(z, r) = B(x, r)$ is automatically contained in our set. Contrast with the archimedean case: in $\mathbb{R}$ with the standard absolute value, the closed interval $[-1, 1]$ is not open — the point $z = 1$ has no open ball of radius $r = 1$ (or any positive radius) contained in $[-1, 1]$. The "every point is a center" property fails because the ultrametric inequality does not hold for the standard absolute value on $\mathbb{R}$. This result also shows that in a non-archimedean field, open and closed balls of the same center and radius are not complementary: $B^\circ(x, r)$ is a proper open subset of $B(x, r)$, and $B(x, r)$ is itself open. In fact, $B(x, r)$ is also closed (being a closed ball by definition), so every closed ball in a non-archimedean field is clopen.[/step]

[guided]We have shown: for every $z \in B(x, r)$, the open ball $B^\circ(z, r)$ satisfies $B^\circ(z, r) \subseteq B(x, r)$. To conclude $B(x, r)$ is open in the metric topology, we verify the definition: a set $S$ is open if for every $z \in S$ there exists $\varepsilon > 0$ such that $B^\circ(z, \varepsilon) \subseteq S$. We have exhibited $\varepsilon = r$ for every $z \in B(x, r)$, so the condition is satisfied. Equivalently, $B(x, r) = \bigcup_{z \in B(x, r)} B^\circ(z, r)$, which is a union of open sets (each $B^\circ(z, r)$ is open by definition of the metric topology), hence open. The strikingly clean choice $\varepsilon = r$ — the same for every interior point — is not a coincidence. In a standard metric space one would typically need $\varepsilon = r - d(x, z) > 0$, which shrinks as $z$ approaches the boundary. In a non-archimedean field there is no "boundary thinning" effect because $B(z, r) = B(x, r)$ for every $z$ already inside: every point is as good a center as any other. This rigidity is what makes non-archimedean geometry so different from Euclidean intuition.[/guided]