If $I_{n+1}\subset I_n$, show that $\bigcap_{n=1}^\infty I_n$ is nonempty Question: If $I_n$ is closed and bounded, $I_{n+1}\subset I_n$, and $I_n\neq\emptyset$, show that $\bigcap_{n=1}^\infty I_n$ is nonempty.
This is not a homework help question. I'm actually looking for a specific proof which does not 1) appeal to the Heine-Borel theorem and 2) use the notions of convergence or sequences.
I recall seeing such a proof before, but I can't remember where.
Addenda: We know $I_n\subset\mathbb R^k$. Either we know $I_n$ is compact or that it is closed and bounded, but not both (I forget the exact initial given information, but I'd appreciate a solution from either direction).
 A: Since this is a decreasing sequence of sets, we might as well treat the ambient space as $I_1$. So we are in a compact space and we have a decreasing sequence of non-empty closed sets.
But now we remember the following characterization of compactness:

If $\{A_i\mid i\in I\}$ is a family of closed sets, such that the intersection of every finite collection of $A_i$'s is non-empty, then $\bigcap_{i\in I}A_i\neq\varnothing$.

So it suffices to see that for every finite collection of $I_n$'s, they have a non-empty intersection. But that's easy because the intersection of $I_{i_1}\cap\ldots\cap I_{i_n}$ is just $I_k$ for $k=\max\{i_j\mid j\leq n\}$. Since $I_k$ is non-empty, we have the wanted property.
By compactness we have that $\bigcap I_n\neq\varnothing$.
A: This proof is taken from baby Rudin, p. 38. Here, $\{I_n\}$ is a sequence of intervals in $\mathbb{R}$. 
" If $I_n = [a_n, b_n]$, let $E$ be the set of all $a_n$. Then $E$ is nonempty and bounded above (by $b_1$). Let $x$ be the sup of $E$. If $m$ and $n$ are positive integers, then $$a_n \leq a_{m+n} \leq b_{m+n} \leq b_m,$$ so that $x\leq b_m$ for each $m$. Since it is obvious that $a_m\leq x$, we see that $x\in I_m$ for each $m = 1,2,3,\ldots.$"
A: Let $X = I_1$, which is compact.
Let $U_n = I_1-I_n $. The $U_n$ are open and nested. Suppose $\displaystyle \cap_{n=1}^\infty I_n$ is empty. Then the $U_n$ cover $I_1$, so by compactness, there exists a finite subcover $U_{n_j}$. Then $$\displaystyle \bigcup_{j=1}^mI_1 -I_{n_j} =I_1,$$ so  $$\bigcap_{j=1}^mI_{n_j}=I_{n_m}=\emptyset$$ - a contradiction
