Proving Delta Sifting Distributionally Starting with the delta distribution: 
$ \langle\delta, \varphi\rangle = \varphi(0)$, where $\varphi \in D(\varphi)$ is a test function
This establishes the "sifting property" for test functions but I want to understand how we can extend this to all continuous functions, $f \in C[\mathbb{R}]$ not necessarily a smooth function, as represented in Dirac notation as $\int_R \delta(x)f(x) dx = f(0) $.
So the question is, how does one arrive at the same result using distributions?
 A: Any distribution $u$ with compact support can be extended to all of $C^\infty$. This is done by taking $\rho \in C_c^\infty$ such that $\rho \equiv 1$ on a neighborhood of the support of $u$ and for $\phi \in C^\infty$ setting
$$\langle u, \varphi \rangle := \langle u, \rho\varphi \rangle.$$
This is well-defined since two such $\rho$ will only differ outside of the support of $u$.

EDIT
How can we apply $\delta$ on a function $\varphi$ that is only continuous?
Let $\psi \in C_c^\infty$ be a mollifier, i.e. $\psi_\epsilon(x) := \epsilon^{-1}\psi(x/\epsilon) \to \delta(x)$ as $\epsilon\to 0$.
Now,
$$
\int_{-\infty}^{\infty} \psi_\epsilon(x) \, \varphi(x) \, dx
= \int_{-\infty}^{\infty} \epsilon^{-1} \psi(x/\epsilon) \, \varphi(x) \, dx
= \{ x=\epsilon y \}
= \int_{-\infty}^{\infty} \psi(y) \, \varphi(\epsilon y) \, dy \\
\to \int_{-\infty}^{\infty} \psi(y) \, \varphi(0) \, dy
= \int_{-\infty}^{\infty} \psi(y) \, dy \, \varphi(0)
= \varphi(0).
$$
Thus we can define $\langle \delta, \varphi \rangle := \lim_{\epsilon\to 0} \int \psi_\epsilon(x) \, \varphi(x) \, dx = \varphi(0).$
A: As @md2perpe notes, every compactly-supported distribution on $\mathbb R^n$ extends to an element of the dual space of the space of smooth functions on $\mathbb R^n$.
In fact, if we pay attention to the Frechet space (projective limit) topology on the space of all smooth functions, we can show that the dual is exactly compactly-supported distributions.
But this seems not to be exactly the question you're asking. You appear to be asking how to legitimately and uniquely extend Dirac $\delta$, arising as an element of the dual space to test functions, to an element of the dual space to the Frechet space of continuous functions $C^0(\mathbb R^n)$. And, indeed, again looking at the Frechet space (projective limit) topology on $C^\infty(\mathbb R^n)$, and using a definition of order of a distribution, we find that all $0$-order compactly-supported distributions give (continuous) linear functionals on $C^0(\mathbb R^n)$.
