Why is the following true about compact spaces? Here's the problem

Let $X$ be a topological space. A family $(V_i)_{i∈I}$ of subsets of $X$ is said to have the finite
  intersection property if for all finite $J ⊆ I$, the intersection $\cap
_{j∈J}
V_j$ is nonempty.
  Prove that $X$ is compact if and only if it has the following property: for every family
  $(V_i)_{i∈I}$ of closed subsets with the finite intersection property, $\cap
_{i∈I}
V_i$
  is nonempty

Now, please note I am not a big fan of compactness. I'm trying to get used to it. That being said, I came up with an odd conclusion.
My understanding of a compact space is that

$X$ is compact if for every open covering, there is a finite subcovering.

And an open covering is just a union of open subsets of $X$.
So, well, if I assume that every family of "closed" subsets of $X$ has the finite intersection property... I concluded that $X$ can't have a bunch of open coverings.
Because, say $V_i,V_j$ are distinct closed subsets of $X$. By the said property, the intersection is non empty. Equivalently, they share at least one element of $X$. Equivalently, $(X＼V_i) \cup (X＼V_j)$ will have a "hole" which is the part that $V_i,V_j$ have in common. So, since every open subset of $X$ must be in the form $X＼V_i$, the union of them will always have a "hole" because the closed subsets are guaranteed to have some element in common, whichever and how many I choose to intersect.
And this "hole" there means that $X$ is not covered how ever many I gather $X＼V_i$s around. That one hole cannot be covered by open subsets. Therefore, even before considering finite subcoverings, $X$ does not have an open covering.
Well, clearly my conclusion "should" be wrong. Problem, is, where and why. I drew a picture with just $2$ closed subsets of $X$ in $X$ but indeed I get a hole in the picture.
Can someone tell me what is wrong with my reasoning?
 A: $X$ is compact i.e for every open covering, $(U_i)_{i\in I}$, there exists a finite subset $J\subset I$ such that $(U_j)_{j\in J}$ cover $X$. Set $C_i$ is the complementary of $U_i$ you obtain:
$X$ is compact if and only if given a family of closed subsets $(C_i)_{i\in I}$ of $X$ such that $\cap_iC_i=\phi$, there exists a finite subset $J\subset I$ such that $\cap_{j\in J}C_j=\phi$.
You know that $p$ implies $q$ is equivalent to the negation of $q$ implies the negation of $p$ so
The contraposition of this definition is for every family $(V_i)_{i\in I}$ if you have for every finite subset $J\subset I$ $\cap_{j\in J}C_j\neq \phi$ then $\cap_{i\in I}C_i\neq \phi$. 
A: What's wrong with your reasoning is that you are starting from the assumption that every family of closed subsets of $X$ has the finite intersection property and drawing some correct but (to you) surprising conclusions from that assumption. The problem is asking you to assume that any family of closed subsets that happens to have the finite intersection property has a non-empty intersection, not that every family of closed sets actually does have the finite intersection property.
[PS: persevere! compactness is really cool!]
A: Suppose that $X$ is not compact, then there is an open cover $\{V_i\}_{i\in I}$ of $X$ such that no finite subcollection covers $X$.
Let $F_i=X\setminus V_i$, then $F_i$ is closed, and each finite intersection
$$ F_{i_1}\cap\dots\cap F_{i_n}=X\setminus (V_{i_1}\cup\cdots\cup V_{i_n})$$
is non-empty because $V_{i_1}\cup\dots V_{i_n}$ doesn't cover $X$. But
$$ \bigcap_{i\in I}F_i=X\setminus\bigcup_{i\in I}V_i=\emptyset $$
because the $V_i$ cover $X$. So $X$ does not have the finite intersection property.
The converse works the same way.
A: You are right that for two intersecting closed sets, their complements are not a cover of $X$. Note that we can reverse this to: if we have a finite open cover of $X$, their complements form a finite family of closed sets with empty intersection. So far so good.
Now what compactness means is that for every open cover, we can find a finite subfamily that is also an open cover. So we have to show that for every family of closed sets with empty intersection (which is not every family of closed sets, of course!) there is a finite subfamily which also has an empty intersection. This follows by the same complements reasoning.
So if $X$ is compact, we also know that last fact about families of closed sets: if we happen to have an empty intersection then this means some finite empty intersection of a subset. So if we assume the latter never happens (which we call the finite intersection property), the former also never happens. This is the basic idea.
