A smooth graph in $\mathbb{R^2}$ is a set of the form $\{(x, f(x))\mid x\in \mathbb{R}\}$ or $\{(f(y), y)\mid y\in \mathbb{R}\}$ for some $f:\mathbb{R} \rightarrow \mathbb{R}$ is some smooth parametrized curve. Show that there is a finite number of graphs $\lambda_1,\dots, \lambda_m$ such that $C_\gamma \subset \bigcup_{i=1}^m \lambda_i$.
$\gamma:[a,b] \rightarrow \mathbb{R^2}$ is a smooth parametrized curve $\iff$ $\gamma$ is $C^1$, $D_+\gamma(a) \neq 0$, $D_-\gamma(b)\neq 0$, $D\gamma(c)\neq 0\ \forall c\in (a,b)$.
$\gamma:[a,b] \rightarrow \mathbb{R^2}$ is a smooth parametrized curve $\implies$ $\gamma$ is a piecewise smooth parameterized curve $\iff$ $\exists$ partition $P=\{ a=a_1<\dots<a_k=b \}$ of $[a,b]$ for which each $\gamma|_[a_i,a_{i+1}]$ are smooth parametrized curves.
$C_\gamma$ is the image of $\gamma$ in $\mathbb{R^2}$, and is connected.
I was thinking that I needed to somehow show that the image of $\gamma$, $C_\gamma$, was of the form that the smooth graphs are of.
You're right. Here's the idea: there are some places $t$ where $\gamma'(t)$ has $0$ as its $y$-component. Near that $t$-value, the graph of $\gamma$ cannot be part of a $y$-axis graph (i.e., $(f(y), y)$. Similarly, near places where $\gamma'(t)$ has an $x$-component zero, it can't be part of an $x$-axis graph. You need to break it into pieces on which one condition or the other does not occur. Then on each piece, the graph of $\gamma$ looks like a function-graph, by the implict function theorem.
Example: $\gamma(t) = (\cos t, \sin t)$ traces out a circle, which is contained in the graphs of $y = \pm \sqrt{1 - x^2}$ and $x = \pm \sqrt{1 - y^2}$.