1. Yes, this is true. This can be derived from the involutive property of the Legendre-Fenchel transform: if $f:\mathbb R^n\to (-\infty,+\infty]$ is a closed convex function, then $f^{**}=f$ (Theorem 12.2 in Rockafellar's Convex Analysis). Recall that
$$f^*(x) = \sup\{\langle x,y\rangle - f(y) : y\in\mathbb R^n\}$$
and a convex function is closed if its epigraph is.
With every convex closed set $K$ we can associate a closed convex function $\delta_K$ by letting $\delta_K(x)=0$ when $x\in K$ and $\delta_K(x)=+\infty$ otherwise. Observe that $\delta_K^*$ is exactly $h_K$.
Given $h$ as in your question, define $K= \{x : \langle x,y\rangle \le h(y) \ \forall y\}$. Observe that $h^*(x)=0$ when $x\in K$, because the supremum is attained by $y=0$. If $x\notin K$, then there is $y$ such that $\langle x,y\rangle > h(y)$. Considering large multiples of such $y$, we conclude that $h^*(x)=\infty$.
Thus, $h^* = \delta_K$. By the involutive property, $h=h^{**} = \delta_K^* = h_K$.
By the way, this result is Theorem 13.2 in Rockafellar's book.
2. Yes, this is correct. Another way to state this fact: the epigraph of $h_{K_1\cap K_2}$ is the convex hull of $\operatorname{epi} h_{K_1} \cup \operatorname{epi} h_{K_2}$. To see this, observe that the epigraph of $h_K$ is determined by its intersection with the horizontal plane $z=1$. This intersection is nothing but $K^\circ$, the polar of $K$. It remains to use the fact that $(K_1\cap K_2)^\circ = \operatorname{conv}(K_1^\circ \cup K_2^\circ)$. See also Corollary 16.5.1 in Rockafellar's book.
If $(K_i)_{i \in \mathbb{N}}$ is a sequence of closed sets in $\mathbb{R}^3$, then the union of these sets $\bigcup_{i=1}^\infty K_i = K_1 \cup K_2 \cup ... $ is also closed.
The above statement is, as you suggest, false. Let $(K_i)_{i \in \mathbb{N}}$ be a sequence of closed sets in $\mathbb{R^n}$.
Then
$\cup_{i=1}^\infty K_i$ is closed iff $(\cup_{i=1}^\infty K_i)^C$ is open, that is iff $\cap_{i=1}^\infty K_i^C$ is open.
Note that each $U_i:=K_i^C$ is open, and as you suggest, you could choose the $U_i$'s so that their intersection is not open, e.g. $U_i = \{v \in \mathbb{R^n} \mid \|v\| < \frac{1}{i}\}$, $\cap_{i=1}^\infty U_i=\{0\}$.
As others have suggested, you could alternatively give a counterexample for the original statement by choosing e.g. $K_i=\{v \in \mathbb{R}^n \mid \|v\| \leq 1-\frac{1}{i}\}$ so that $\cup_{i=1}^\infty K_i=\{v \in \mathbb{R}^n \mid \|v\| < 1\}$.
P.S. In your question you used the same symbol $K_i$ to stand for two different notions (a closed and an open set). This is a very bad practice which you should avoid as it confuses the reader.
Best Answer
The claim is obviously true for $k=1$. It is not true for $k\ge2$:
Define the sets $$ S:= \{ x\in \mathbb R^k: \|x\|_2\le 1, \ x_1=0\}, $$ which is a closed, $k-1$-dimensional ball, and $$ B_n = \{x\in \mathbb R^k: \|x\|_2\le 1, \ x_k \le 1- \frac1n\}, $$ which is a ball that is cut. Then $$ K_n = conv(S \cup B_n) $$ is an increasing sequence of convex and compact sets. And its union $K=\cup K_n$ is the closed unit ball:
Let $\|x\|_2\le 1$. If $x_k=1$ then $x\in S \subset K_n$, if $x_k<1$ then $x\in B_n$ for all $n$ such that $1-\frac1n\ge x_k$, or equivalently for all $n$ with $n\ge \frac1{1-x_k}$.