If a compact supported distribution $T\in\cal D'(\mathbb{R}^n)$ has support $K$, we all know that for any smooth function $u\in C^\infty_c(K^c)$, $\langle T,u\rangle=0$. My question is that is it true that for any $u\in C^\infty_c(\mathbb{R}^n)$ with $\mathrm{supp}(u)\subset \overline{K^c}$, which means $\mathrm{supp}(u)\cap K$ may not be empty, and we also have $\langle T,u\rangle=0$?
I think if we can find a sequence of smooth functions $u_j\in C^\infty_c(K^c)$, s.t. $u_j\to u (D(\mathbb{R}^n))$, thus we have $0=\langle T,u_j \rangle \to \langle T,u \rangle$. But I haven't found proper sequence to approximate $u$.
My answer inpired by Steele:
Define $$\xi(x) = \begin{cases}1 \text{ if }K+O_{2\varepsilon,}\\ 0 \text{ else,} \end{cases}$$ and let $\xi_{\varepsilon}=\xi*\alpha_{\varepsilon}$ where $\alpha_{\varepsilon}$ is the smooth kernel. Then $\xi_{\varepsilon} = 1$ if $x\in K+O_{2\varepsilon}$ and $\xi_{\varepsilon} = 0$ if $x\in (K+O_{3\varepsilon})^c$. And $|\partial^\alpha\xi_\varepsilon|<C \varepsilon^{-|\alpha|}$.
Since $T\in\mathcal{D}'$, there exiets $C,m$, s.t. $$|<T,\phi>|<C\sup\limits_{|\alpha|<m,x\in\mathbb{R}^n} |\partial^\alpha\phi(x)|\ \ \ \ \ \ \forall\phi\in C^\infty_c(K+O_1).$$ Thus, if $\varepsilon<1/2$, we obtain $$|<T,u>|=|<T,\xi_\varepsilon u>|<C\sup\limits_{|\alpha|<m,x\in\mathbb{R}^n} |\partial^\alpha(\xi_\varepsilon u(x))|=\mathcal{O}(\varepsilon).$$ So, we get $<T,u>=0$ where $u\in C^\infty_c(\mathbb{R}^n)$ with $\mathrm{supp}(u)\subset \overline{K^c}$.(The last equation hold on because $\partial^\alpha u(x)=0 \text{ if } x\in\partial{K}$)
I assume that you meant that $u\in C^\infty_0(K^c)$ that is $u$ and all its derivatives vanish on $K$, as $C^\infty_c(\overline{K^c})$ was meaningless ($\overline{K^c}$ can be the whole of $\Bbb{R}^n$).
Take $\psi\in C^\infty_c(|x|<1), \int\psi = 1$, $\psi_m(x)=m^n \psi(mx)$ and $$f_m = 1_{dist(x,K)<2/m} \ast \psi_m$$
Due to the polynomial growth of $\|\partial^\alpha f_m\|_\infty$ and the faster-than-polynomial decay of $\partial^\alpha u(x)$ in $dist(x,K)$ you'll get that $$\forall \alpha, \lim_{m\to \infty} \| \partial^\alpha (f_m u) \|_\infty=0$$ ie. $f_m u\to 0$ in $C^\infty_c(\Bbb{R}^n)$ and $$\lim_{m\to \infty} \langle T ,f_m u\rangle = 0 $$
As $\langle T,u\rangle=\langle T,f_m u\rangle$ we get your result that $$\langle T, u\rangle=0$$