Let's start at the bottom and work our way up.
Why is $(\mu I - T)$ continuous?
The topology considered on $D_T$ is transported from the subspace topology on the graph inherited from the product topology on $X\times X$. The graph $\Gamma(T) = \{ (x,Tx) : x \in D_T\}$ is a linear subspace of $X\times X$, and any norm $\lVert\,\cdot\,\rVert_{X\times X}$ that induces the product topology on $X\times X$ yields a norm on $D_T$ that defines the considered topology via $\lVert x\rVert_{D_T} := \lVert (x,Tx)\rVert_{X\times X}$. Common choices of the norm on $X\times X$ are $(x,y) \mapsto \bigl(\lVert x\rVert_X^p + \lVert y\rVert_X^p\bigr)^{1/p}$ for $1 \leqslant p < +\infty$ or $(x,y) \mapsto \max \{ \lVert x\rVert_X, \lVert y\rVert_X\}$. For all these choices it is clear that $\lVert x\rVert_X \leqslant \lVert x\rVert_{D_T}$ for all $x\in D_T$, so the topology induced by $\lVert\,\cdot\,\rVert_{D_T}$ on $D_T$ is finer than the subspace topology. Since by assumption $T$ is closed, i.e. $\Gamma(T)$ is a closed subspace of $X\times X$, and hence a Banach space under the norm $\lVert\,\cdot\,\rVert_{X\times X}$, $D_T$ is a Banach space under $\lVert\,\cdot\,\rVert_{D_T}$, and hence the topology is strictly finer than the subspace topology unless $D_T$ is closed with respect to $\lVert\,\cdot\,\rVert_X$, in which case $T$ is continuous with respect to $\lVert\,\cdot\,\rVert_X$ and the two topologies on $D_T$ coincide.
Now $\lVert x\rVert_X \leqslant \lVert x\rVert_{D_T}$ means that the inclusion $i = I\lvert_{D_T}$ is continuous (with norm $\leqslant 1$; but the norm depends on the choice of $\lVert\,\cdot\,\rVert_{X\times X}$, while the continuity doesn't). For the above choices of the norm on $X\times X$, it is also easy to see that $\lVert Tx\rVert_X \leqslant \lVert x\rVert_{D_T}$, so $T \colon D_T \to X$ is continuous (again, with norm $\leqslant 1$ for the above choices of the norm on $X\times X$). And since linear combinations of continuous linear maps are continuous, it follows that $(\mu I - T) \colon D_T \to X$ is continuous for all $\mu \in \mathbb{C}$.
Why does continuity of $(\mu I - T)^{-1}$ and (a) imply that $R_T(\mu) = i(\mu I - T)^{-1}$ is compact?
The composition of a continuous operator and a compact operator is compact. So compactness of $i$ and continuity of $(\mu I - T)^{-1}$ together imply compactness of their composition $R_T(\mu)$.
To cover the cases where the compact operator is the first to be applied and the cases where it's the second with one argument, consider the composition of three operators, the outer ones continuous and the middle one compact. The case of the composition of two operators is obtained by letting one of the outer maps be the identity. Hence, let $E,F,G,H$ Banach spaces, and $A\colon E \to F$ continuous, $B \colon F \to G$ compact, $C \colon G \to H$ continuous (all linear of course). Then $C \circ B \circ A$ is compact.
I don't know which definition of compact operators you use, so I'll give the argument for the two most common characterisations.
A linear operator $K \colon Y \to Z$ is compact if there is a neighbourhood $U$ of $0$ in $Y$ such that $K(U)$ is relatively compact in $Z$ (this definition also works for topological vector spaces that aren't Banach spaces, but the argument that follows needs a small modification for the general case). For Banach spaces, this is easily seen to be equivalent to "the image of every bounded set is relatively compact". So let $U$ be the unit ball in $E$. By continuity of $A$, the image $A(U)$ is bounded. By compactness of $B$, therefore $\overline{B(A(U))}$ is compact, and by continuity of $C$ it follows that $C\bigl(\overline{B(A(U))}\bigr)$ is compact, whence $$(C\circ B\circ A)(U) = C\bigl(B(A(U))\bigr) \subset C\bigl(\overline{B(A(U))}\bigr)$$ is relatively compact, i.e. $C\circ B \circ A$ is compact.
A linear operator $K \colon Y \to Z$, where $Y$ and $Z$ are Banach spaces, is compact if every bounded sequence $(y_n)$ in $Y$ has a subsequence $(y_{n_k})$ such that $\bigl(K y_{n_k}\bigr)$ is convergent. Then let $(y_n)$ a bounded sequence in $E$. Since $A$ is a bounded operator, $Ay_n$ is a bounded sequence in $F$. Since $B$ is compact, we can extract a subsequence $(y_{n_k})$ such that $B(Ay_{n_k})$ is convergent in $G$, and since $C$ is continuous, the sequence $C(B(Ay_{n_k}))$ is also convergent. Thus $C\circ B \circ A$ is compact.
Why is $(\mu I - T)^{-1} \colon X \to D_T$ continuous?
That follows from the closed graph theorem.
Since $R_T(\mu) \colon X \to X$ is continuous, its graph $\Gamma$ is a closed subspace of $X\times X$. Since the image of $R_T(\mu)$ is $D_T$, we can view $\Gamma$ as a subspace of $X\times D_T$. If we endow $X\times D_T$ with the subspace topology inherited from $X\times X$, then $\Gamma$ is closed in $X\times D_T$. The topology of $D_T$ is finer than the subspace topology $D_T$ inherits from $X$, so the product topology on $X\times D_T$ is finer than the subspace topology it inherits from $X\times X$. Hence $\Gamma$ is closed as a subspace of $X\times D_T$ when we endow that with the product topology. But $\Gamma$ is the graph of $(\mu I - T)^{-1}$, and $D_T$ and $X$ are Banach spaces, so the closed graph theorem yields the continuity of $(\mu I - T)^{-1} \colon X \to D_T$.
