Suppose $(X,d)$ is an unbounded metric space. Show that there exists a sequence in $X$ that has no convergent subsequence.
My thoughts:
The space is unbounded so there is a sequence $x_n$ in $X$ that has no Cauchy subsequence. That means that there is a subsequence $x_{n_k}$ that does not converges.
Pick $x_0 \in X$. As $X$ is unbounded, $X$ does not equal $B(x_0, 1)$, so we can pick $x_1\notin B(x_0,1)$. Next we pick $x_2 \notin B(x_0,1)\cup B(x_1,1)$ ( the latter finite union is bounded so cannot equal $X$).
We continue by recursion and construct a sequence $(x_n)$ (by for each $n$ picking $x_{n+1} \notin \bigcup_{i=0}^n B(x_i,1)$) such that all mutual distances are at least $1$ (if $x_n, x_m$ and $n>m$ we chose $x_n \notin B(x_m,1)$ in particular), and such a sequence cannot have a convergent (or even Cauchy) subsequence.