[proofplan]
We first verify that the supremum norm is a non-archimedean norm on $C(\mathbb{Z}_p, \mathbb{Q}_p)$ by checking non-degeneracy, multiplicativity of the underlying field norm, and the strong triangle inequality. For completeness, we show that every Cauchy sequence $(f_n)$ converges uniformly: the pointwise limits define a function $f$, and uniform convergence (from the Cauchy condition) ensures $f$ is continuous and $\|f_n - f\| \to 0$.
[/proofplan]
[step:Verify the supremum norm is a non-archimedean norm]
Define $\|f\| := \sup_{x \in \mathbb{Z}_p} |f(x)|_p$ for $f \in C(\mathbb{Z}_p, \mathbb{Q}_p)$. Since $\mathbb{Z}_p$ is compact and $f$ is continuous, this supremum is attained, so $\|f\| = \max_{x \in \mathbb{Z}_p} |f(x)|_p \in \mathbb{R}_{\geq 0}$.
**Non-degeneracy:** $\|f\| = 0$ iff $|f(x)|_p = 0$ for all $x \in \mathbb{Z}_p$ iff $f(x) = 0$ for all $x$ iff $f = 0$.
**Homogeneity:** For $\lambda \in \mathbb{Q}_p$, $\|\lambda f\| = \sup_x |\lambda f(x)|_p = |\lambda|_p \sup_x |f(x)|_p = |\lambda|_p \|f\|$.
**Strong triangle inequality:** For $f, g \in C(\mathbb{Z}_p, \mathbb{Q}_p)$ and any $x \in \mathbb{Z}_p$,
\begin{align*}
|f(x) + g(x)|_p \leq \max(|f(x)|_p, |g(x)|_p) \leq \max(\|f\|, \|g\|),
\end{align*}
where the first inequality uses the non-archimedean property of $|\cdot|_p$ on $\mathbb{Q}_p$. Taking the supremum over $x$ gives $\|f + g\| \leq \max(\|f\|, \|g\|)$.
[/step]
[step:Show every Cauchy sequence in $C(\mathbb{Z}_p, \mathbb{Q}_p)$ converges pointwise to a well-defined function]
Let $(f_n)_{n \geq 1}$ be a Cauchy sequence in $(C(\mathbb{Z}_p, \mathbb{Q}_p), \|\cdot\|)$. For each fixed $x \in \mathbb{Z}_p$:
\begin{align*}
|f_m(x) - f_n(x)|_p \leq \|f_m - f_n\| \to 0 \quad \text{as } m, n \to \infty.
\end{align*}
So $(f_n(x))_{n \geq 1}$ is a Cauchy sequence in $\mathbb{Q}_p$. Since $\mathbb{Q}_p$ is complete, the limit exists. Define
\begin{align*}
f: \mathbb{Z}_p &\to \mathbb{Q}_p \\
x &\mapsto \lim_{n \to \infty} f_n(x).
\end{align*}
[/step]
[step:Prove $f_n \to f$ uniformly and $f$ is continuous]
Given $\varepsilon > 0$, choose $N$ such that $\|f_m - f_n\| < \varepsilon$ for all $m, n \geq N$. Then for every $x \in \mathbb{Z}_p$ and $m, n \geq N$:
\begin{align*}
|f_m(x) - f_n(x)|_p \leq \|f_m - f_n\| < \varepsilon.
\end{align*}
Fixing $n \geq N$ and letting $m \to \infty$, the continuity of $|\cdot|_p$ gives $|f(x) - f_n(x)|_p \leq \varepsilon$ for all $x \in \mathbb{Z}_p$. Taking the supremum over $x$: $\|f - f_n\| \leq \varepsilon$ for all $n \geq N$. This shows $f_n \to f$ uniformly.
Since each $f_n$ is continuous and the convergence is uniform, $f$ is continuous: for any $x_0 \in \mathbb{Z}_p$ and $\varepsilon > 0$, choose $n$ with $\|f - f_n\| < \varepsilon/2$, then choose $\delta > 0$ with $|f_n(x) - f_n(x_0)|_p < \varepsilon/2$ for $|x - x_0|_p < \delta$. The strong triangle inequality gives
\begin{align*}
|f(x) - f(x_0)|_p \leq \max(|f(x) - f_n(x)|_p, |f_n(x) - f_n(x_0)|_p, |f_n(x_0) - f(x_0)|_p) < \varepsilon.
\end{align*}
So $f \in C(\mathbb{Z}_p, \mathbb{Q}_p)$ and $\|f_n - f\| \to 0$, completing the proof that $C(\mathbb{Z}_p, \mathbb{Q}_p)$ is a Banach space.
[guided]
The completeness argument follows the standard pattern for supremum-norm spaces, but with the non-archimedean norm replacing the usual triangle inequality.
Given a Cauchy sequence $(f_n)$ in $C(\mathbb{Z}_p, \mathbb{Q}_p)$, we need to produce a limit $f$ that is (i) a function from $\mathbb{Z}_p$ to $\mathbb{Q}_p$, (ii) continuous, and (iii) the limit of $(f_n)$ in the supremum norm.
**Step (i)** is straightforward: the pointwise limits exist because $\mathbb{Q}_p$ is complete and each sequence $(f_n(x))$ is Cauchy (since $|f_m(x) - f_n(x)|_p \leq \|f_m - f_n\|$, the supremum norm bound passes to each point).
**Step (iii)** is the key uniformity argument. The Cauchy condition $\|f_m - f_n\| < \varepsilon$ for $m, n \geq N$ says the oscillation $|f_m(x) - f_n(x)|_p$ is uniformly small in $x$. Sending $m \to \infty$ with $n$ fixed, the non-archimedean absolute value is continuous (as a function $\mathbb{Q}_p \to \mathbb{R}$), so $|f(x) - f_n(x)|_p = \lim_{m \to \infty} |f_m(x) - f_n(x)|_p \leq \varepsilon$ for each $x$. The bound is independent of $x$, so $\|f - f_n\| \leq \varepsilon$.
**Step (ii)** follows from (iii): $f$ is a uniform limit of continuous functions. The $\varepsilon/3$-argument works verbatim, with the non-archimedean $\max$ replacing the archimedean sum (and in fact giving a slightly cleaner estimate since $\max(\varepsilon/2, \varepsilon/2) = \varepsilon/2 < \varepsilon$).
[/guided]
[/step]
