How do I show that a contraction mapping in a metric space is continuous? I start out by letting $V$ be an arbitrary open set in $X$.  Then 
$$
f^{-1}(V) = \{x\in X\mid f(x) \in B_\epsilon(f(a))\}.
$$
This can be re-written as:
$$
f^{-1}(V) = \{x\in X\mid d(f(a), f(x)) < \epsilon \}.
$$  I realize that contraction mappings have an $0<r<1$ such that 
$$
d(f(x_1), f(x_2)) \leq r\cdot d(x_1,x_2),\quad \forall x_1,x_2 \in X.
$$  I construct an open ball 
$$
B_{\frac{\delta}{r}}(a) = \{x\in X\mid r\cdot d(a, x) \lt \delta \}
$$ but from here I'm unsure as to how to show that $f^{-1}(V)$ is open.

Let $(X, d)$ be a metric space, $S \subset X$, and $f:S \longrightarrow S$ a function be such that $d(f(x), f(y)) \leq c d(x, y)$, for all $x, y \in S$, where $0 \leq c < 1$ is given.
Fix $\epsilon > 0$ and choose $a \in S$. 
The case where $c = 0$ is trivial. Assume $c > 0$ and let $\delta = \epsilon / c$.
For all $x \in S$ with $d(x, a) < \delta$, we have $d(f(x), f(a)) \leq c d(x, a) < \epsilon$, i.e. $f$ is continuous.
