Let $X$ be a metric space. Suppose that for some $\epsilon > 0$, every $\epsilon$-ball in $X$ has compact closure. Show that $X$ is complete.
I'd like to solve the above problem. For $X$ to be complete, every Cauchy sequence should converge. A Cauchy sequence could be comprised with elements in a $\epsilon$-ball of X since those are all has distance smaller than $\epsilon$.
But how could I show every those comprised sequence being converged to where with the hint "compact" & "closure"?
With compact I can imagine those $\epsilon$-balls might be covered with finite open elements, and with closure, there always exists a limit point...but can't be sure that existence of a limit point guarantees the convergence of a Cauchy sequence. How will those limit points and compact will make the sequence convergent?
I'm not certain I follow your reason exactly, but here's how I would approach it.
Suppose $(x_n)_{n=1}^\infty$ is a Cauchy sequence. By Cauchiness, there exists an $N \in \mathbb{N}$ such that $$m, n \ge N \implies \|x_n - x_m\| < \varepsilon.$$ Therefore, for $n \ge N$, we have $\|x_n - x_N\| < \varepsilon$, hence $x_n \in B(x_N; \varepsilon)$ for $n \ge N$. Since $\overline{B(x_N; \varepsilon)}$ (note: in general this is not necessarily $B[x_N; \varepsilon]$!) is compact, the subsequence $(x_n)_{n=N}^\infty$ contains a convergent (sub-)subsequence $(x_{n_k})_{k=1}^\infty$. Since $(x_n)_{n=1}^\infty$ is Cauchy with a convergent subsequence $(x_{n_k})_{k=1}^\infty$, we have that $(x_n)_{n=1}^\infty$ is convergent.