Suppose that we have a function $\psi(x)$ so that
$$
\delta(\phi)=\int_{\mathbb{R}}\psi(x)\phi(x)\,\mathrm{d}x\tag{1}
$$
Define
$$
\begin{align}
\eta(x)&=\left\{\begin{array}{}
0&\text{if }|x|\le\frac12\\
\frac{\displaystyle e^{\frac1{1-2|x|}}}{\displaystyle e^{\frac1{1-2|x|}}+e^{\frac1{|x|-1}}} &\text{if }\frac12\lt|x|\lt1\\
1&\text{if }|x|\ge1
\end{array}\right.\tag{2a}\\[6pt]
\eta_k(x)&=\eta\!\left(2^{-k} x\right)\tag{2b}
\end{align}
$$
Then $\eta_k\in C^\infty$, $\eta_k(x)=0$ for $|x|\le2^{k-1}$, and $\eta_k(x)=1$ for $|x|\ge2^k$. Since $C^\infty$ is dense in $L_\text{loc}^1$, for any $\epsilon>0$, we can choose a $\psi_\epsilon\in C^\infty$ so that
$$
\begin{align}
\|(\psi-\psi_\epsilon)(1-\eta)\|_{L^1}&\le\epsilon/2\tag{3a}\\
\|(\psi-\psi_\epsilon)(\eta_k-\eta_{k+1})\|_{L^1}&\le2^{-k-2}\epsilon\quad\text{for }k\ge0\tag{3b}
\end{align}
$$
which means
$$
\|\psi-\psi_\epsilon\|_{L^1}\le\epsilon\tag{3c}
$$
Furthermore, let $\sigma$ be the signum function, then since $\psi_\epsilon\in L_\text{loc}^\infty$, we can choose a $\sigma_\epsilon\in C^\infty$ so that $|\sigma_\epsilon|\le1$ and
$$
\begin{align}
\|(\sigma\circ\psi-\sigma_\epsilon)(1-\eta)\psi_\epsilon\|_{L^1}&\le\epsilon/2\tag{4a}\\
\|(\sigma\circ\psi-\sigma_\epsilon)(\eta_k-\eta_{k+1})\psi_\epsilon\|_{L^1}&\le2^{-k-2}\epsilon\quad\text{for }k\ge0\tag{4b}
\end{align}
$$
which means
$$
\|(\sigma\circ\psi-\sigma_\epsilon)\psi_\epsilon\|_{L^1}\le\epsilon\tag{4c}
$$
Since $\|\sigma\circ\psi-\sigma_\epsilon\|_{L^\infty}\le2$, we get
$$
\begin{align}
\|(\sigma\circ\psi-\sigma_\epsilon)\psi\|_{L^1}
&\le\|(\sigma\circ\psi-\sigma_\epsilon)\psi_\epsilon\|_{L^1}+\|(\sigma\circ\psi-\sigma_\epsilon)(\psi-\psi_\epsilon)\|_{L^1}\tag{5a}\\
&\le\|(\sigma\circ\psi-\sigma_\epsilon)\psi_\epsilon\|_{L^1}+\|\sigma\circ\psi-\sigma_\epsilon\|_{L^\infty}\|\psi-\psi_\epsilon\|_{L^1}\tag{5b}\\
&\le3\epsilon\tag{5c}
\end{align}
$$
Define $\phi_k(x)=\sigma_\epsilon(x)\left(\eta(2^{-k} x)-\eta(2^{-k-1} x)\right)$. Then we have
$$
\begin{align}
0
&=\delta\left(\sum_{k=m}^n\phi_k\right)\tag{6a}\\
&=\sum_{k=m}^n\int_{\mathbb{R}}\psi(x)\,\phi_k(x)\,\mathrm{d}x\tag{6b}\\
&=\sum_{k=m}^n\int_{\mathbb{R}}\psi(x)\,\sigma_\epsilon(x)\left(\eta(2^{-k} x)-\eta(2^{-k-1} x)\right)\mathrm{d}x\tag{6c}\\
&=\int_{\mathbb{R}}\psi(x)\,\sigma_\epsilon(x)\left(\eta(2^{-m} x)-\eta(2^{-n-1} x)\right)\mathrm{d}x\tag{6d}\\
&\ge\int_{\mathbb{R}}|\psi(x)|\left(\eta(2^{-m} x)-\eta(2^{-n-1} x)\right)\,\mathrm{d}x\\
&-\int_{\mathbb{R}}|(\sigma\circ\psi(x)-\sigma_\epsilon(x))\psi(x)|\left(\eta(2^{-m} x)-\eta(2^{-n-1} x)\right)\mathrm{d}x\tag{6e}\\
&\ge\int_{2^m\lt|x|\lt2^n}|\psi(x)|\,\mathrm{d}x-3\epsilon\tag{6f}
\end{align}
$$
Explanation:
$\text{(6a)}$: $\phi_k\in C_c^\infty$ and is $0$ in a neighborhood of $0$
$\text{(6b)}$: apply $(1)$
$\text{(6c)}$: apply the definition of $\phi_k$
$\text{(6d)}$: compute the sum of the $\eta_k$
$\text{(6e)}$: $\psi\sigma_\epsilon=|\psi|-(\sigma\circ\psi-\sigma_\epsilon)\psi$
$\text{(6f)}$: apply $(5)$
Since $\epsilon$ was arbitrary, $(6)$ implies that on any annulus centered at $0$, $\psi$ is $0$. That is, $\psi(x)=0$ for all $x\ne0$. Since the value of a function at a single point does not affect the integral of that function, $(1)$ would imply that
$$
\delta(\phi)=0\tag7
$$
for all $\phi$. Since $(7)$ is false, $(1)$ must be false.
One way of making sense of what $\int_a^b f(x) \delta(x) dx$ means is to see the Dirac delta as a measure. Now, I won't go into the details of what a measure is (if you are curious, head to the wikipedia page!). Let's just say that a measure is something that lets us make sense of the integral of real-valued functions over subsets. For a function $f: \mathbb{R} \to \mathbb{R}$, the integral of $f$ over the subset $A \subset \mathbb{R}$ with respect to the measure $\mu$ is usually written either
$
\int_{A} f \; d\mu
$
or
$
\int_{A} f(x) \; d\mu(x)
$
(if one wants to specify the variable of integration). Let me emphasize that it is very important that we specify the subset over which we are integrating!
The usual integral of functions over $\mathbb{R}$ is the integral with respect to Lebesgue measure, and the classical notation for the Lebesgue integral of a function $f$ over an interval with endpoints $a$ and $b$ is
$\int_{a}^{b} f(x) \; dx$. Now, notice that this last notation is ambiguous because it might mean either of the following integrals:
$$
\int_{[a,b)} f(x) \; dx \quad \text{or} \quad
\int_{(a,b)} f(x) \; dx \quad \text{or} \quad
\int_{(a,b]} f(x) \; dx \quad \text{or} \quad
\int_{[a,b]} f(x) \; dx
$$
However, in the case of Lebesgue measure, it turns out that all these integrals are equal and so the notation actually makes sense. The reason behind this is that the Lebesgue measure of a set is essentially its volume. Since points have zero volume, it is safe to ignore them in integrals.
Now, the point is that writing $\int_{a}^b f(x) \delta(x) dx$ is an abuse of notation, which can be dangerous if one is not careful. The Dirac delta is not a function on $\mathbb{R}$, so it does not make sense to integrate it with respect to the Lebesgue measure. What is really meant behind this notation is that $\int_{a}^b f(x) \delta(x) dx$ is the integral of $f$ with respect to the $\delta$ measure, namely it is the integral $\int_A f \; d\delta$, where $A$ is one of the intervals with endpoints $a$ and $b$.
Recall that in the case of Lebesgue measure, in didn't matter what choice of interval $A$ we made. For the Dirac measure, it is crucial that we specify over which interval we integrate. Here is why. Loosely speaking, the measure $\delta$ assigns "volume" $1$ to the set $\{ 0 \}$ and every set that contains $0$, and assigns "volume" $0$ to every other set, i.e. every set that doesn't contain $0$. This means that all the volume of the real line is concentrated at the point $0$. A consequence of this is that the integral of a function is given by:
$$
\int_A f \;d \delta =
\begin{cases}
f(0) & \text{if} \; 0 \in A\\
0 & \text{if} \; 0 \notin A
\end{cases}
$$
For most purposes, it is safe to think of this last equation as the definition of the Dirac delta. Now, notice that with this formula we have , for $a < 0$,
$$
\int_{(a, 0)} f \; d \delta = 0
$$
since $0 \notin (a, 0)$ while
$$
\int_{(a, 0]} f \; d \delta = f(0)
$$
since $0 \in ( a, 0]$. So, while it did not make any difference if we forgot about the point $0$ when computing the Lebesgue integral, it makes a huge difference when integrating with respect to the Dirac measure, because all the volume of the Dirac measure is concentrated at the point $0$.
To sum up, in order to make sense of the integral $\int_a ^b f(x) \delta(x) dx$, you should be careful to indicate what set you are integrating on. The notation $\int_A f(x) \; d\delta(x)$ can be a useful way of doing this.
Now to your next question: can we break up integrals in pieces?
Integrals with respect to measures share many properties with the usual Riemann integral (and in other ways they are very different, as I pointed out earlier). For example, the integral is linear. Another key property is that if a set $A$ can be written as a disjoint union of two sets, that is $A = B \cup C$ where $B \cap C = \emptyset$, then we can break up the integral:
$$
\int_A f \; d\mu = \int_B f \; d\mu + \int_C f \; d\mu.
$$
This means that you can safely break up integrals, as long as you keep track of the sets you are integrating on. What you should avoid at all cost is the following mistake:
$$
\int_a^b f(x) \delta(x) dx = \int_{a} ^0 f(x) \delta(x) dx + \int_0 ^b f(x) \delta(x) dx = f(0) + f(0)
$$
Best Answer
Suppose $d$ is a distribution with compact support $S$ and $f$ is a smooth function, not necessarily with compact support. Define a smooth, compactly supported cutoff function $k$ such that $k=1$ on a neighborhood of $S$. Then $\langle d,kf \rangle$ makes proper sense. One can prove that the value of this expression does not depend on $k$, under the preceding assumptions. So this defines a meaningful extension of $d$ to all of $C^\infty$. Now that we know the key idea is that of a compactly supported distribution, a quick google search finds a proof: http://www.math.chalmers.se/~hasse/distributioner_eng.pdf page 31.
Now the Dirac delta has support $\{ 0 \}$, so this applies to it.
The principle is the same if you want to apply $d$ to a function like $\chi_{[a,b]}$ which is smooth on a neighborhood of the support of $d$ but not globally smooth: use a smooth cutoff which keeps the function the same on a neighborhood of the support of $d$ but such that the result is globally smooth. So this procedure works for $\delta[\chi_{[a,b]}]$ provided $a,b \neq 0$.