Finding example of a complete metric space I'm trying to find a complex metric space. There are several examples that are fairly obvious -- the real numbers with the Euclidean metric and any set with the discrete metric come to mind. I'm trying to prove my conjecture that any closed subset of $\mathbb{R}$ is also a complete space, which follows from the lemma that for any closed subset $F \subset X$ and sequence $\{p_n\}$ in $F$ which converges, say to $p$, we have $p \in F$.
I'm going to show my attempt for $[0,1]$. So, let $\{p_n\}$ be a Cauchy sequence in $[0,1]$. Then $p_n \to p$ in $\mathbb{R}$ since $\mathbb{R}$ is complete, so $\{p_n\}$ is convergent. Call its limit $p$. By the aforementioned lemma, since $[0,1]$ is closed, we have $p \in [0,1]$.
The problem is that this proof relies on completeness of $\mathbb{R}$. Is there a way to prove this that doesn't require it? Or, perhaps better yet, is there a better example of a complete metric space?
 A: Your proof is correct. In fact more is true, instead of $\mathbb{R}$, if you take any complete metric space$(X, d) $ and instead of $ [0, 1]$ , if you take any closed subset $F$ of $X$, then the you can also conclude that $(F, d_F) $ is complete metric space, where $d_F$ is the subspace metric induced by $d$ on $F$.Proof is similar to your proof.
Suppose, $(x_n) $ be a Cauchy sequence in $(F, d_F)$ , this implies $(x_n) $ is also Cauchy sequence in $(X, d) $ . Since, $(X, d) $ is complete, $(x_n)$ converges in $(X, d) $.Since, $F\subset X $ is closed,$\lim (x_n)\in F $

The problem is that this proof relies on completeness of R. Is there a
way to prove this that doesn't require it?

Well, $F= [0,1]$ is closed and bounded subset of $(\mathbb{R},$|$  $ |$)$, hence compact.
Let, $(x_n) $ be a Cauchy sequence in the metric subspace $[0, 1]$, then by compactness $\exists$ a convergent subsequence $(x_{n_k}) $ of $(x_n) $ and it is not difficult to prove that $(x_n) $ converges to $\lim(x_{n_k}) $. But, this doesn't avoid the completeness of $\mathbb{R}$!
Completeness of $(X, d) $ is necessary to prove a closed subset $F\subset X$ is complete.
To justify, take $(X=\mathbb{Q},$d=|$    $|$ ) $
and $F=[2, 3]\cap \mathbb{Q}$ closed subset of $(X,d)$ but not complete. Because the sequence $(x_n)_{n\in {\mathbb{N}}}$ defined by
$x_n=(1+\frac{1}{n}) ^n$ is Cauchy sequence in $(F, d_F) $ but not convergent .
The sequence $(x_n) $ converges to $e$ which is not in $F$, in fact it is not in $X$ also.
Hence, completeness of $X, d) $ is necessary, otherwise any cauchy sequence in $(F, d_F) $ may not have limit in $(X, d) $, ( it may
have the limit in some larger space due to the fact that every metric space has a completion).

Or, perhaps better yet, is there a better example of a complete metric
space?

There are lot of examples of complete metric space.
$(X, d_{discrete}) $ , $(\mathbb{R^n},d_{euclidean}) $, $(C[a, b], d_{\infty}) $ etc.
