The Nested Triangles Theorem (in the complex plane). It is possible? Today I've read a known theorem in Complex Analysis, the so called The Nested Rectangles Theorem. In the first part of this theorem and proof that I've read, is enunciated the classical theorem, and in second part ensure that eventually (all but a finite numbers) all the rectangles lie inside an arbirary small disc.  
I've thought about the possibility of a theorem for triangles, notice that I don't require equilateral triangles. (I say that in previous sequence of rectangles we can find a triangle lie in the first, and including the third, and so on with the second triangle and the next three rectangles, I don't know if this intuituion is good or useful. Too I've thought about the possibility of use barycentric coordinates.)

Question. Can you state and give a proof of The Nested Triangles Theorem?

I believe that it could be a nice and useful exercise. Thanks in advance.
 A: Define a closed triangle in a plane to be a convex hull of three non-collinear points. Note that a closed triangle is a closed set in $\Bbb R^2$. For a closed triangle, define its diameter to be the longest of its sides (which is also a maximal distance between points in this triangle). The the following holds:

If $t_1\supseteq t_2\supseteq t_3\supseteq...$ is a sequence of closed triangles diameters of which tend to zero, then there is precisely one point contained in all of them.

I will prove there is at least one point. Showing that there is at most one follows easily on diameter condition.
Suppose there is no point contained in all of the triangles. Then $t_2^c,t_3^c,...$ (complements of the triangles) form an open cover of $t_1$ (can you see why this is true?). Because by Heine-Borel theorem $t_1$ is compact, this open cover has a finite subcover, let's say it's $t_{n_1}^c,t_{n_2}^c,...,t_{n_k}^c$, where $n_1<n_2<...<n_k$. But $t_{n_1}^c\cup t_{n_2}^c\cup...\cup t_{n_k}^c=t_{n_k}^c$, which means $t_1\subseteq t_{n_k}^c$. But then $t_{n_k}\subseteq t_{n_k}^c$, which is a contradiction, as $t_{n_k}^c$ is non-empty.
Let me just note that this can be generalized to any nested family of closed sets with diameters tending to zero. The existence of a point is an instance of the following more general phenomenon:

A set $X$ in some topological space is compact iff it has finite intersection property, i.e. if the following holds for every family $\mathcal F$ of closed subsets of $X$: If every finite subfamily of $\mathcal F$ has nonempty intersection, then so does whole $\mathcal F$.

After seeing my argument above, it will be instructive exercise to try to prove this equivalence.
A: You can also prove the following statement: in a complete metric space any decreasing sequence$(T_n)$  ($T_n \supset T_{n+1}$ for all $n$) of closed subset with diameters $\textrm{diam}(T_n)\to 0$ has an intersection consisting of a unique point. The uniqueness is easy, since the intersection, if non-void, must have diameter $0$. To show that it is non-void, take a point $x_n \in T_n$ for every $n$. We have 
$d(x_n,x_m) \le \textrm{diam}(T_n)$ for all $n \le m$, so the sequence $(x_n)$ is Cauchy. Let $x = \lim x_n$. Since for all $n_0$ $x$ is the limit of the sequence $(x_n)_{n\ge n_0}$, and the closed set $T_{n_0}$ contains this sequence, we have $x\in T_{n_0}$. This is true for any $n_0$, so $x \in \cap_n T_n$. 
