Prove $[0,1]$ is closed, now I know there are a variety of different ways to show this but I'm trying to stick with the things I was showed so far.
A subset $S \subseteq \mathbb{R}^n$ is complete if every Cauchy sequence $(\bar{x}^{(k)})$ in $S$ converges to an element of $S$.
A subset $S \subseteq \mathbb{R}^n$ is closed if whenever $(\bar{x}^{(k)})$ is a sequence in $S$ such that $(\bar{x}^{(k)}) \rightarrow \bar{x}$, then $\bar{x} \in S$.
And $S \subseteq \mathbb{R}^n$ is complete $\iff$ $S$ is closed.
So I see two ways of proceeding, show $[0,1]$ is complete or complete. For the former, show that for any Cauchy sequence consisting of elements in $[0,1]$, converges to an element of $S$. Or that an sequence in our set converges to something in $[0,1]$.
Idea: Elements of any sequence of S are bounded between 0 and 1, that is $0 \leq x_i \leq 1$, where $x_i$ is part of some sequence, show somehow that it means that it converges to something $0 \leq x \leq 1$. Use Bolzano-Weirstrass because we know the sequence is bounded?
Would appreciate help.
Let $(x_{k})$ be a sequence in $[0,1]$ and assume that $x_{k}\to x$. We will show that $x\in [0,1]$. Assume on contrary that $x\not\in[0,1]$. We have two cases:
Case one: $x>1$. In this case, consider the midpoint $y=\frac{x+1}{2}$ of $1$ and $x$. Since $1<y<x$ and since each $x_{k}$ lies in $[0,1]$, it must be that $x_{k}<y$ for all $k$. Thus, letting $\epsilon:=x-y>0$, we see that for each $k$, $x-x_{k}>x-y=\epsilon$. Thus contradicts that $x_{k}\to x$. (Otherwise, we should have $x-x_{k}<\epsilon$ for sufficiently large $k$.)
Case two: $x<0$. I'll leave this to you. It is very similar to case one.