Define $\Sigma=\{0,1\}^\mathbb{N}$, the infinite sequences of $0s$ and $1s$, with the metric $$d(\sigma,\sigma')=\frac{1}{\min\{k:a_k\neq b_k\}}$$ where $\sigma=(a_k)_1^\infty,\sigma'=(b_k)_1^\infty.$
I am looking to show that this is a complete metric space, and since I am very new to metric spaces, I just wanted to check that my approach is correct.
$\mathbf{My\; attempt:}$
Let $(\sigma_n)\subset\Sigma$ be a Cauchy sequence.
For any $M\in\mathbb{N}$, there is $N\in\mathbb{N}$ such that $$n,m\geq N\implies d(\sigma_n,\sigma_m)\lt\frac{1}{M}\iff \min\{k:\sigma_n,\sigma_m \text{ differ}\}\gt M.$$ Fix $m$ and varying $n\geq N$, we see $\sigma_n$ has the same first $M$ terms (determined by $\sigma_m$) for any $n\geq N$, and $M$ was arbitrary so we see that we can construct a sequence $\sigma$ from these terms.
Then for any $\epsilon\gt0$ choose $M\gt\frac{1}{\epsilon}$, so that there is $N$ such that $n\geq N$ gives $$\min\{k:\sigma,\sigma_n\text{ differ}\}\gt M\iff d(\sigma,\sigma_n)\lt\frac{1}{M}\lt\epsilon.$$
$\square$
My main aim is for this to be fully rigorous but I am unsure how to rigorously construct the limit, as it feels slightly handwavy! Any advice would be great, thanks.
Consider a Cauchy sequence in $\Sigma$, that is a sequence of sequence $\left(\left({u^m}_n\right)_{n\in\mathbb{N}} \right)_{m\in\mathbb{N}}$. What you have shown is that for a fixed $n \in \mathbb{N}$, the sequence $m \mapsto {u^m}_n$ is eventually constant: it follows that it converges to a value ${u^{\infty}}_n$.
Show that $\left({u^{\infty}}_{n}\right)_{n\in\mathbb{N}}$ is the desired limit of the Cauchy sequence $\left(\left({u^m}_n\right)_{n\in\mathbb{N}} \right)_{m\in\mathbb{N}}$.