Fix $p \in B((0, 0, 1), 1)$ and denote the line joining $(0, 0, 2)$ and $p$ by $L_p$. Let $(a, b, 0)$ be the intersection of $L$ and the $xy$-plane, then
$L_p = \{(0, 0, 2) + t[(a, b, 0) - (0, 0, 2)] \mid t \in \mathbb{R}\} = \{(ta, tb, 2 - 2t) \mid t \in \mathbb{R}\}.$
By assumption, $L$ intersects the ball $B((0, 0, 1), 1)$ so it intersects the sphere $\partial B((0, 0, 0), 1)$ twice. Therefore there are two real solutions to $(ta)^2 + (tb)^2 + (2 - 2t)^2 = 1$ which simplifies to $(a^2 + b^2 + 4)t^2 - 8t + 3 = 0$. In order for this quadratic to have two real solutions, we must have
\begin{align*} 0 &< \Delta\\ &= (-8)^2 - 4(a^2 + b^2 + 4)(3)\\ &= 64 - 12a^2 -12b^2 - 48\\ &= 16 - 12a^2 - 12b^2\\ &= 4(4 - 3(a^2+b^2)). \end{align*}
Therefore, $a^2 + b^2 < \frac{4}{3}$. Conversely, if $a^2 + b^2 < \frac{4}{3}$, then the line joining $(0, 0, 2)$ and $(a, b, 0)$ intersects the ball $B((0, 0, 1), 1)$. So
$D = \bigcup_{p \in B((0, 0, 1), 1)}L_p = \bigcup_{a^2 + b^2 < \frac{4}{3}} L_{(a, b, 0)}.$
Let $(x, y, z) \in D$, then $(x, y, z) = (at, bt, 2 - 2t)$ for some $a, b, t \in \mathbb{R}$ with $a^2 + b^2 < \frac{4}{3}$. If $t = 0$, $(x, y, z) = (0, 0, 2)$. If, $t \neq 0$, that is $(x, y, z) \in D\setminus\{(0, 0, 2\}$, we have
$x^2 + y^2 = a^2t^2 + b^2t^2 = (a^2 + b^2)t^2 < \frac{4}{3}t^2 = \frac{4}{3}\left(1 - \frac{1}{2}z\right)^2 = \frac{1}{3}(2 - z)^2.$
Conversely, given $(x, y, z) \in \mathbb{R}^3$ such that $x^2 + y^2 < \frac{1}{3}(2 - z)^2$, note that $2 - z\neq 0$, so
\begin{align*} \left(\frac{x}{2 - z}\right)^2 + \left(\frac{y}{2-z}\right)^2 &< \frac{1}{3}\\ \left(\frac{2x}{2 - z}\right)^2 + \left(\frac{2y}{2-z}\right)^2 &< \frac{4}{3}. \end{align*}
Furthermore, $(x, y, z) = (x_0t, y_0t, 2 - 2t)$ for $t = \frac{1}{2}(2 - z)\neq 0$ where $x_0 = \dfrac{2x}{2 - z}$ and $y_0 = \dfrac{2y}{2-z}$. So $(x, y, z) \in D\setminus\{(0, 0, 2)\}$.
Therefore,
$D = \left\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 < \frac{1}{3}(2-z)^2\right\}\cup\{(0, 0, 2)\}.$
That is, $D$ is the point $(0, 0, 2)$ together with the set of points enclosed by the (double) cone defined by $x^2 + y^2 < \frac{4}{3}(2-z)^2$, but not including the points on the cone itself.
With this description of the set $D$ at hand, the problem becomes much simpler. I will answer the second question first as it is easier. The closure, interior and boundary of $D$ are
\begin{align*} \overline{D} &= \left\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 \leq \frac{1}{3}(2-z)^2\right\}\\ D^{\circ} &= \left\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 < \frac{1}{3}(2-z)^2\right\}\\ \partial D &= \left\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 = \frac{1}{3}(2-z)^2\right\}. \end{align*}
As $D \neq \overline{D}$, $D$ is not closed and as $D^{\circ} \neq D$, $D$ is not open.
You can give a more complete proof of the fact that that $D$ is neither open nor closed, but once you understand what $D$ is geometrically, I think it is pretty clear.