Munkres Topology, page 102, questions 19:c and 19:d PROBLEM

If $A \subset X$, we define the boundary of $A$ by the equation
  $$\text{Bd } A = \bar{A} \cap \overline{X - A}.$$

[Munkres Topology, page 102, question 19:c] Show that
    $$U \text{ is open } \iff \text{Bd } U = \bar{U} - U.$$
[Munkres Topology, page 102, question 19:d] If $U$ is open, is it true that $U = \text{Int }(\bar{U})$?  Justify your answer.


MY ATTEMPT
I am pretty sure I must somehow use the results in either Munkres Topology, page 102, question 19:a or Munkres Topology, page 102, question 19:b, but I am having trouble seeing how.
Sure, I can think of examples where both
$$U \text{ is open } \iff \text{Bd } U = \bar{U} - U$$
and
$$U \text{ is open } \implies U = \text{Int }(\bar{U})$$
hold.
(It suffices to take either $U=\mathbb{R}$ or $U=\emptyset$ under the usual topology.)
Risking looking dumb in this forum, let me try to prove [19:c]:
Suppose that $U$ is open. By a result in this answer, $\overline U = \text{Int } U$.  But $\overline{U} = \text{Int } U \cup \text{Bd } U$ (by Munkres Topology, page 102, question 19:a).  Now, I am stuck here, and I honestly do not know how to go further.
Lastly, I think the solution for [19:d] must somehow use some result(s) from [19:c], but again I am not sure how.
QUESTIONS


(1) How do you show [19:c]?
(2) What is a counterexample (if any) for [19:d]?  Otherwise, how do we go about proving it?


 A: For (1), suppose $U$ is open. If $x$ belongs to $\operatorname{Bd}(U)$, you want to show that $x$ is a point of the closure of $U$, but not a point of $U$ itself. Since every neighborhood of $x$ contains points from $U$ and $U^c$, $x$ cannot be an interior point of $U$, so $x$ is a point of the closure of $U$, but not a point of $U$ itself.
For the other inclusion, if $x\in\overline U-U$, then every neighborhood of $x$ contains a point of $U$ since $x\in\overline U$. Since $x$ is not a point of $U$, it is not an interior point of $U$. I'll let you finish that one.
Now, if $\operatorname{Bd}(U) = \overline U - U = \overline U\cap U^c$, we claim $U$ is open. We will do this by showing that $U^c$ is closed. Suppose that $x$ is a limit point of $U^c$. If $x\in U$, then every neighborhood of $x$ contains a point from $U^c$ and a point from $U$, so that $x$ is a boundary point of $U$. But then $x$ belongs to $U^c\cap U$, which is impossible, so the claim is proved.
For (2), think about the sets $(0,1)$ and $(1,2)$ as subsets of $\Bbb R$ with the standard topology.
