Let be $(C([0,1]),\Vert \cdot\Vert_{\infty})$ the normed space of continuous functions, equipped with the supremum norm $\Vert\cdot\Vert=\sup\limits_{x\in[0,1]}|f(x)|$.
Show that $B_1(0):=\{f\in C([0,1])\mid\Vert f\Vert_{\infty}\leq 1\}$ is a closed set.
Let be $(f_n)_{n\in\mathbb{N}}$ a convergent sequence (of functions) and $f$ its limit (function). Then, for an arbitrary $\epsilon>0$ we find an $n_0\in\mathbb{N}$ such that for all $n>n_0$ we have that
$$\begin{align*} &\Vert f\Vert_{\infty}-\Vert f_n\Vert_{\infty}\leq \Vert f-f_n\Vert_{\infty}<\epsilon\\ &\implies \Vert f\Vert_{\infty}<\epsilon+\Vert f_n\Vert_{\infty}\leq\epsilon+1. \end{align*}$$ As $\epsilon>0$ is arbitrary it must be $\Vert f\Vert_{\infty}\leq 1$ because otherwise $\Vert f\Vert_{\infty}$ was not the smallest upper bound. Hence, $f\in B_1(0)$.
I was 100% sure that this is correct but my tutor left a comment that I can't do it this way and that I should review the sample solution. Where is my mistake?