Claim: A topological space $\,X\,$ is compact iff it has the Finite Intersection Property (=FIP):
Proof: (1) Suppose $\,X\,$ is compact and let $\,\{V_i\}\,$ be a family of closed subsets s.t. $\,\displaystyle{\bigcap_{i}V_i=\emptyset}\,$. Putting now $\,A_i:=X-V_i\,$ , we get that $\,\{A_i\}\,$ is a family of open subsets , and
$$\bigcap_{i}V_i=\emptyset\Longrightarrow \;X=X-\emptyset=X-\left(\bigcap_iV_i\right)=\bigcup_i\left(X-V_i\right)=\bigcup_iA_i\Longrightarrow\;\{A_i\}$$
is an open cover of $\,X\,$ and thus there exists a finite subcover of it:
$$X=\bigcup_{i\in I\,,\,|I|<\aleph_0}A_i=\bigcup_{i\in I\,,\,|I|<\aleph_0}(X-V_i)=X-\left(\bigcap_{i\in I\,,\,|I|<\aleph_0}V_i\right)\Longrightarrow \bigcap_{i\in I\,,\,|I|<\aleph_0}V_i=\emptyset\Longrightarrow$$
The family $\,\{V_i\}\,$ has the FIP.
(2) Suppose now that every family of closed subsets of $\,X\,$ hast the FIP, and let $\,\{A_i\}\,$ be an open cover of it. Put $\,U_i:=X-A_i\,$ , so $\,U_i\, $ is closed for every $\,i\,$:
$$\bigcap_iU_i=\bigcap_i(X-A_i)=X-\bigcup_i A_i=X-X=\emptyset$$
By assumption, there exists a finite set $\,J\,$ s.t. $\,\displaystyle{\bigcap_{i\in J}U_i=\emptyset}\,$ , but then
$$X=X-\emptyset=X-\bigcap_{i\in J}U_i=\bigcup_{i\in J}X-U_i)=\bigcup_{i\in J}A_i\Longrightarrow$$
$\,\{A_i\}_{i\in J}$ is a finite subcover for $\,X\,$ and thus it is compact....QED.
Please be sure you can follow the above and justify all steps. Check where we used Morgan Rules, for example, and note that we used the contrapositive of the FIP's definition...
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
$$V_1\cap \bar{V_2}\cap V_3\cap\cdots = \bar{V_1}\cap\bar{V_2}\cap\bar{V_3}\cap\cdots.$$
Best Answer
Suppose $x \in S_1$. As we assumed at the start that $\bigcap_{n=1}^\infty S_n =\emptyset$, $x$ cannot be a member of all $S_n$ so some $n_0$ exists such that $x \notin S_{n_0}$, or equivalently $x \in V_{n_0}$. So the $V_n$ cover $S_1$.
Note that as the $S_n$ are decreasing, so the $V_n$ are increasing, so when we have a finite subcover, the one with the largest index (which exists by finiteness) is a superset of all of them, so we have a subcover consisting of one set, say $V_k$. But this cannot be, as any point of $S_k$ (which exists, as these sets are non-empty) is by definition not covered by $V_k$. This contradiction shows that the original assumption of $\bigcap_{n=1}^\infty S_n = \emptyset$ was false.