The open ball $B(f,1)$ in $(C[0,1],\mathbb R), d_{\infty})$ compact? Which of the following are compact?


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*The open ball $B(f,1)$ in $(C[0,1],\mathbb R), d_\infty)$, where $f:[0,1] \to \mathbb R$ denotes the constant function at zero and $d_\infty (f,g) = \sup_{x\in[0,1]} |f(x)-g(x)|$


My guess is that this is not compact because $(C[0,1],\mathbb R), d_\infty)$ is unbounded.


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*The closed ball $D(0,1)$ in $(\mathbb Z,d)$, where $d$ denotes the 3-adic metric.


My guess is that this is not compact, because $(\mathbb Z,d)$ is not complete.

Are either of these correct? Could anyone provide better proofs of them?
 A: The open ball $B(f,1)$ in $(C[0,1],\mathbb R), d_\infty)$, where $f:[0,1] \to \mathbb R$ denotes the constant function at zero and $d_\infty (f,g) = \sup_{x\in[0,1]} |f(x)-g(x)|$ is clearly bounded, since every point in it is within distance $1$ from just one point. No two points in it are as much as, nor more than $2$ units away from each other, by the triangle inequality.
But the space is not compact, as may be seen as follows. Let $g_n$ be a continuous function that is $0$ outside the interval whose endpoints are $n\pm\frac 1 3$ has the value $1$ at $n$. That sequence has no convergent subsequence.
A: Well, my guess is that:$\\$
if $f(x)\neq g(x)$ for every $x\Rightarrow d_\infty(f,g)>0 \Rightarrow \exists$ $B(f(x),\frac{d_\infty}{2}), B(g(x),\frac{d_\infty}{2})\\ $, where $B(f(x),\frac{d_\infty}{2}) \cap B(g(x),\frac{d_\infty}{2})=\emptyset $, so this topological space is a Hausdorff Space, and since a compact in a Hausdorff space must be closed, an open ball cannot be a compact.
Proof that a compact subspace in a Hausdorff space is closed may be found in Serge Lang's Analysis book at page 32.
