(1) is easier to asnwer: If $X$ is a metric space and $Y \subset X$ is a dense subset, then $\overline Y = X$. So a proper closed subspace cannot contain a dense subspace.
(2) is just a definition, it does not say that $D(A)$ is closed. It says that $A$ is called closed if and only if the graph $\{(x, Ax): x\in D(A)\}$ is closed. $D(A)$ is Not closed.
Let's have an example. Let $A : D(A) \to L^2(\mathbb R)$ be defined by
$$Af (x) = xf(x),$$
and $D(A) = \{ f\in L^2(\mathbb R): xf \in L^2(\mathbb R)\}$. Note that $D(A)$ is dense in $L^2(\mathbb R)$ as it contains all continuous function with compact support.
CLaim: $A$ is closed. Let $\phi_n \to \phi$ and $x\phi_n \to \psi$ in $L^2(\mathbb R)$. Then by passing to a subsequence if necessary, we can assume the convergence are almost everywhere. Thus we have $x\phi(x) = \psi(x)$ almost everywhere. In particular, we have $\phi \in D(A)$ and $A\phi = \psi$. Thus $A$ is closed.
Now $A$ is densely defined and closed, but $D(A)$ is not closed: the function
$$f = \min \{1, 1/|x|\} \in L^2(\mathbb R)$$
is not in $D(A)$.
(If you consider for example $f_n = \chi_{[-n, n]} f$. Then $f_n \to f$ in $L^2(\mathbb R)$. But $Af_n \in L^2(\mathbb R)$ does not converge)
Best Answer
No. For example, consider the Laplacian with Dirichlet boundary conditions on $(0,1)$. This operator has a bounded inverse by Poincaré's inequality and $X=C_c^\infty(0,1)$ is contained in $D(T)$ and dense in $L^2(0,1)$. However, if $f\in C_c^\infty(0,1)$, then $$ \int_0^1 \Delta f\,dx=f'(1)-f'(0)=0, $$ so that constant functions are orthogonal to $T(X)$. Hence $T(X)$ cannot be dense in $L^2(0,1)$.