Show that $S^1$ is complete Define: $S^1 = \{(x,y): x^2+y^2 = 1\}$

Then wish to show $S^1$ is complete

Attempt:

Let $(z_n)$ be a Cauchy sequence on $S^1$, $(z_n) = (x_n, y_n)$
Then $S^1$ is complete if $\forall \epsilon > 0, \exists N,  d(z_n,z_m) = \sqrt{(x_n - x_m)^2 + (y_n - y_m)^2}< \epsilon, $for $n, m \geq N$
However, $\sqrt{(x_n - x_m)^2 + (y_n - y_m)^2} = 1 \Rightarrow (x_n - x_m)^2 + (y_n - y_m)^2 = 1$
So it is not true that $d(z_n, z_m) < \epsilon $ for $n, m \geq N$...

But we know $S^1$ is complete since it is a compact subset of $\mathbb{R}^2$. Can someone spot and correct my mistake? 
I have never done a Cauchy sequence proof with more than one variable, how to proceed?

What you have to show is that any Cauchy sequence in $S^1$ has a limit that belongs to $S^1$. What you did is not correct because it's not true that the distance from two points in $S^1$ is always $1$. For example, take the points $z_1 = (1,0)$ and $z_2 = (0,1)$. 
To show this, you start with a Cauchy sequence $\{z_n\}_{n\in \mathbb{N}} \in S^1$. Writing $z_n = (x_n,y_n)$, we know that the sequences $\{x_n\}_{n\in \mathbb{N}}$ and $ \{y_n\}_{n\in \mathbb{N}}$ are Cauchy sequences of real numbers, so they must converge to some reals $x$ and $y$, because $\mathbb{R}$ is complete. 
What we have to do now is see if $z=(x,y) \in S^1$. This is easy because we know that the distance is continuous, so $d(z,0) = \displaystyle\lim_{n \rightarrow \infty} d(z_n,0) = 1$, because $d(z_n,0) =1$ for every $n \in \mathbb{N}$
