As I understand it, two analytic functions defined on $\mathbb{R}^k$ which agree on a set with an accumulation point must agree on a neighborhood; however, the same is not true of $C^\infty$ functions. What is an example of two $C^\infty$ functions which agree on a set containing an accumulation point, but do not agree on any neighborhood?
(The functions $f\equiv 0$ and $g(x) = \begin{cases} 0 &\text{ if $x\leq0$}\\ e^{\frac{-1}{x}}&\text{ if $x > 0$}\end{cases}$ agree on a set which has zero as an accumulation point, but do not agree in a neighborhood of zero. However, they do agree on some open set.)
njguliyev's example: $0$ and $e^{-1/x^2} \sin\frac{1}{x}$ is probably the best that one can give. But I'll mention a somewhat related result (due to Whitney, I guess): for every closet subset $A\subset \mathbb R$ there is a $C^\infty$ function $f:\mathbb R\to\mathbb R$ such that $f^{-1}(0)=A$. For example, you can choose $A$ to be the Cantor set and get a $C^\infty$ function that agrees with $0$ on the Cantor set and only there.