Must the intersection of a certain decreasing sequence of open sets be non-empty I am trying to show the open mapping theorem. As I was trying to prove it, I made the following conjecture:

Let $X$ be a complete metric space. Let $V_1,V_2,...$ be a sequence of
  open sets in $X$ such that: $$V_1\supseteq \overline{V_2}\supseteq
 V_2\supseteq \overline{V_3}\supseteq V_3\supseteq
 \overline{V_4}\supseteq V_4\supseteq .... $$Then
  $\cap_{n=1}^{\infty}V_n\not=\emptyset$

Question 1: Is the above conjecture true ?
I would be able to prove the open mapping theorem if the conjecture is true, but I am curious about this question as well:
Question 2: Does the conjecture hold even if we don't require $X$ to be complete ?
Thank you

Edit: Neal answered both questions. There is still one more question that Ii formulated after seeing the counterexamples of Neal.
Question 3: Does the conjecture hold if $X$ is a complete bounded metric space ?
 A: No, the conjecture need not be true. Consider $V_i = (i,\infty)\subset\mathbb{R}$. Then:
$$(1,\infty)\supseteq [2,\infty)\supseteq (3,\infty)\supseteq \cdots$$
but (by the Archimedean principle) their intersection is empty.
If $X$ is bounded but not complete, the conjecture still does not hold: Consider the punctured interval $X = [-1,0)\cup(0,1]$ and a $$V_i = \bigg(-\frac{1}{i},\frac{1}{i}\bigg)\cap X.$$

If $X$ is just bounded and complete, I think it won't work. We'll take a disjoint union of countably many little balls and then let $V_i$ be the union of all but finitely many of them.
Consider a separable infinite-dimensional Hilbert space with orthonormal basis $e_i$. Define auxiliary sets 
$$E_{i,\epsilon} = B(e_i,\epsilon).$$ 
Choose $\epsilon$ small enough so that the $E_{i,\epsilon}$ are pairwise disjoint and put
$$V_i = \bigcup_{j=i}^\infty E_{j,\epsilon/i}$$
Now for each $i$, we've $\bar{V_i}\subseteq V_{i-1}$, and yet the intersection of all $V_i$ is empty.
Since $V_1$ is contained in the ball of radius $1 + \epsilon$, it is a subset of the bounded, closed metric space $\overline{B(0,1+\epsilon)}$. 
This example may be easier to visualize if you consider the simpler space defined as a countably infinite collection of intervals $[0,1]$ wedged at $0$, i.e.,
$$\bigsqcup_{i\in\mathbb{N}} [0,1]_i/(\forall i,j\ \ 0_i\sim 0_j).$$
Let $E_{(j,\epsilon)} = (1-\epsilon, 1]_j$, the open $\epsilon$-neighborhood of the endpoint of the $j^{th}$ interval, and $V_i$ the union of all but the first $i$ of the $E_{j,\epsilon/i}$s. As $i$ increases, the size of each $E_j$ shrinks, so $\overline{V_{i+1}}\subseteq V_i$ for all $i$. But any $x\in V_1$ is only contained in finitely many $\overline{V_i}$, so the intersection is empty.

However, if $X$ is compact, you can get your result by arguing that
$$V_1\cap \bar{V_2}\cap V_3\cap\cdots = \bar{V_1}\cap\bar{V_2}\cap\bar{V_3}\cap\cdots.$$
A: Question 1 : false .  take $X = \mathbb{R}$, $V_n = (-\infty, -n)$.
