Infinite dimensional Banach spaces and sequences

Question

Let $X$ be an infinite dimensional Banach space.
$a)$ Prove that there is a sequence $x=(x_n)_{n \geq 1}$ in $X$ such that $||x_n||=1$ for all $n$ satisfying
\begin{equation*} \mathrm{dist}(x_{n+1},\mathrm{Span} \{x_1,...,x_n \})\geq 1, n\geq 1 \end{equation*} $b)$ Based on $a)$, show that the closed unit ball in $X$ cannot be compact.

My attempt:
I want to prove it by induction: if $n=0$, then I can take any $0 \neq x \in X$ and normalize it. Thus $\mathrm{dist}(x,\mathrm{Span} \{ 0 \})= ||x||=1$ and the inequality is true. Suppose now it's true for $n-1$. Then for $x_1,...,x_n$ I can choose $x_{n+1} \notin \mathrm{Span}\{x_1,...,x_n\}$ (since $X$ is infinite dimensional) such that $||x_{n+1}||=1$. Thus $\mathrm{dist}(x_{n+1},\mathrm{Span} \{x_1,...,x_n \})=\mathrm{inf}_{x \in \mathrm{Span}\{x_1,...,x_n\} }d(x_{n+1},x) \geq 1$ since $d(x_{n+1},x)=d(x,x_{n+1}) \geq \mathrm{dist}(x, <x_{n+1}>) \geq 1$ ($<x_{n+1}>$ is $1$ dimensional and I can apply the induction hypothesis). In particular I get that $||x_n-x_m|| \geq 1$ for $n \neq m$
For part $b)$ suppose the closed unit ball is compact, i.e. every sequence in it has a convergent subsequence. But this is not true for my sequence $(x_n)_{n \geq1}$ since $||x_n-x_m|| \geq 1$ for $n \neq m$. Am I right?