A symmetric operator $T$ is called essentially self-adjoint if its closure $T$ is self-adjoint. If $T$ is closed, a subset $D \subset D(T)$ is called a core for $T$ if $\overline {T\upharpoonleft D} = T$. So any dense subset of $D(T)$ is a core for $T$.
I would like to know about if there are any other equivalent conditions for a subset $D \subset D(T)$ to be a core for $T$.
On
Yes, there is a nice check:
A subset $\mathcal{D}\subseteq\mathcal{D}(A)$ is a core for a closed operator $A$ iff the following condition is satisfied: $$(x_0,Ax_0)\perp(x,A\restriction_\mathcal{D}x)\text{ for all }x\in\mathcal{D}\implies x_0=0$$
That condition basically comes from the orthogonal decomposition for closed subspaces: $$\mathcal{G}(A)=\overline{\mathcal{G}(A\restriction_\mathcal{D})}\oplus\mathcal{G}(A\restriction_\mathcal{D})^\perp$$ But one has to be aware that the operator must have been closed so that: $$A\text{ closed}\implies\mathcal{G}(A)\text{ closed}\implies\mathcal{G}(A)\text{ complete}$$
And yes, there is the ready core:
For a closed and densely defined operator a core is given by: $\mathcal{D}(A^*A)$
That one basically comes while using the above from a subtle investigation of: $(1+A^*A)$
No, it's not true that any dense subset of $D(T)$ is a core for $T$. For $D \subseteq D(T)$ to be a core for $T$, what you want is $\{(x,Tx): x \in D\}$ to be dense in the graph $\{(x,Tx): x \in D(T)\}$ of $T$ (as subsets of ${\mathscr H} \times {\mathscr H}$, with the norm topology).
Equivalently, for any $x \in D(T)$ there exists a sequence $x_n \in D$ such that $\|x_n - x\| + \|T x_n - Tx\| \to 0$ as $n \to \infty$.