Topological space in which every filter converges to every point Let $(X, \mathcal{T})$ be a topological space in which every filter converges to every element $x \in X$. Show that $\mathcal{T}$ is the trivial topology.
I'm kind of stuck on this one for a while now. It is pretty easy to see that in a set X, endowed with the trivial topology, every filter converges to every element $x \in X$; but I'm not able to prove the proposition stated above. I'm really looking for a constructive proof, not a proof by contraposition. All my attempts fail to say something about the open parts of $X$ if this makes any sense.
Definitions used (for the sake of consistency):


*

*A filter is a nonempty set $\mathcal{F}$ for which the following properties hold: $\mathcal{F}$ does not contain the empty set; for every  $F \in \mathcal{F}$ such that $F \subset G, G \in \mathcal{F}$ holds; for every $F \in \mathcal{F}$ and $G \in \mathcal{F} $ also $F \cap G \in \mathcal{F}$.

*A filter $\mathcal{F}$ converges to $x \in X$ iff the neighbourhood filter of $x$ is contained in $\mathcal{F}$ or, equivalently, for every $V$ in the neighbourhood filter of $x$, there exists an element $F \in \mathcal{F}$ such that $F \subset V$.

*A subset $V \subset X$ is called a neighbourhood of $x$ if there exists an open set $T \in \mathcal{T}$ such that $x \in T \subset V$.

*The neighbourhood filter of $x$ is the set of all neighbourhoods of $x$.

 A: Let $\mathcal F$ be the filter $\{X\}$.  If $V$ is any nonempty open subset of $X$, choose any $y \in V$.  Then $V$ is a member of the neighborhood filter of $y$, hence $V \in \mathcal F$ by hypothesis.  Hence $V = X$.
A: EDIT: This answer is wrong, but I think it is a good example of a wrong proof for this problem. For a good proof, you can approach the problem the way I did but, instead of working with an arbitrary filter, look at the filter {X} (as said by @D_S) and then the result will follow directly.
Take $A \in \mathcal{T} \backslash \{X \cup \phi\}$. Take an arbitrary filter $\mathcal{F}$ and take $x \in A$ and $y \in X\backslash A$. 
Since $\mathcal{F}$ converges to both x and y, there exist $F$ and $G$ in $\mathcal{F}$ such that $F \subset V$ and $G \subset W$ for $V$ and $W$ elements of the neighbourhood filter of $x$ and $y$ respectively.
Further, we know that $A$ is an element of the neighbourhoodfilter of $x$ and $int(X\backslash A)$ is an element of the neighbourhoodfilter of $y$.
The definition of a filter then yields that $\phi = A \cap (X\backslash A) \in \mathcal{F}$, which is clearly not possible. Thus, $\mathcal{T} = \{\phi, X\}$.
