Let $M$ be a smooth manifold, and $C^\infty_0(M)$ the space of smooth compactly supported functions on $M$. Let $\mathcal{D}'(M)$ be the space of distributions, meaning continuous linear functionals $\mathfrak{J} : C^\infty_0(M)\to \mathbb{R}$.
Now, we have the following definitions:
Definition 1. Let $U\subset M$, we say that $\mathfrak{J}$ vanishes in $U$ if for every $\phi \in C^\infty_0(M)$ with $\operatorname{supp}\phi\subset U$ we have $\mathfrak{J}[\phi]=0$.
Definition 2. Define $$\mathfrak{U}=\{U\subset M : \text{$U$ is open and $\mathfrak{J}$ vanishes on $U$}\},$$
we define the support of $\mathfrak{J}$ to be $$\operatorname{supp}\mathfrak{J}=M\setminus \bigcup \mathfrak{U}.$$
i.e., as the complement of the union of all open sets on which $\mathfrak{J}$ vanishes.
Now consider the following situation: $\mathfrak{J}$ depends on the values of the test functions just on $A\subset M$. This means that if $\phi_1,\phi_2\in C^\infty_0(M)$ with $\phi_1|_A = \phi_2|_A$ then $\mathfrak{J}[\phi_1]=\mathfrak{J}[\phi_2]$.
We can thus prove the following claim:
Proposition: If $A$ is closed, $\operatorname{supp} \mathfrak{J}\subset A.$
Proof: To show that $\operatorname{supp}\mathfrak{J}\subset A$ it is enough to prove that $M\setminus A\subset \bigcup \mathfrak{U}$.
Since $A$ is closed, $M\setminus A$ is open. On the other hand $\mathfrak{J}$ vanishes on $M\setminus A$. Indeed, let $\phi\in C^\infty_0(M)$ have $\operatorname{supp}\phi\subset A$, meaning that $\phi|_A =0$. Thus $\mathfrak{J}[\phi]=\mathfrak{J}[0]$ and the later is zero by linearity, showing that $\mathfrak{J}$ vanishes in $A$.
This shows that $M\setminus A\in \mathfrak{U}$ and hence $M\setminus A\subset \bigcup \mathfrak{U}$ which in turn shows that $\operatorname{supp}\mathfrak{J}\subset A$.
Now, I've used that $A$ is closed and hence $M\setminus A$ is open in order to conclude $M\setminus A\in \mathfrak{U}$ (since all sets on this collection are open), which seems one important part of the proof.
But what if $A$ is not closed? It seems reasonable to me that whatever $A$ is if $\mathfrak{J}[\phi]$ depends on $\phi$ just inside $A$ we should have $\operatorname{supp}\mathfrak{J}\subset A$. Is that indeed the case? How the proof would change?
The difference is subtle but important. In the case of smooth functions, we define $\operatorname{supp} \eta$ to be the closure of $A = \{ x \in M : \eta(x) \neq 0 \}$. Of course $\eta$ is "determined by its values in $A$", but $\operatorname{supp} \eta \not\subseteq A$.
To see that this is the case also for distributions, let $M = \mathbb R$ and choose $\mathfrak{J}$ to be $$ \mathfrak{J}[\phi] = \int_{\mathbb R} \eta(x) \phi(x) \, dx, $$ where $\eta \in C_c^\infty(\mathbb R)$ is a fixed non-zero function. Then clearly $\mathfrak J$ only depends on the values of the test functions on $A$, but $\operatorname{supp} \eta \not\subseteq A$.
In general, the proof you presented shows that $\operatorname{supp} \eta \not\subseteq \overline{A}$, but not more.