Intersection of sequence of compact non empty sets has a similar diameter to the elements in the sequence? Suppose $(K_n)$ is a sequence of nested, compact, nonempty sets, where $K_1 \supset K_2 \supset K_3 \supset \cdots,$ and $K = \bigcap_{n \in \mathbb{N}} K_n.$ If for some $x > 0,$ $\operatorname{diam}K_n \geq x$ for all $n \in \mathbb{N},$ is it true that $\operatorname{diam}K \geq x.$
Suppose, for contradiction, that $\operatorname{diam}K < x.$ That is, $\nexists p,q \in K$ such that $d(p,q) \geq x.$ 
In other words, $\nexists p,q \in \bigcap_{n \in \mathbb{N}} K_n$ such that $d(p,q) \geq x.$ 
So, for some $n \in \mathbb{N},$ $\nexists p,q \in K_n$ such that $d(p,q) \geq x,$ contradicting the given assumption.
Hence, $\operatorname{diam}K \geq x.$
 A: For each $n$, pick $p_n,q_n \in K_n$ such that $ d(p_n, q_n) \ge x$. Now take a common convergent subsequence of each. 
A: No, your proof is flawed.

Your last line

So, for some $n \in \mathbb{N},$ $\nexists p,q \in K_n$ such that $d(p,q) \geq x$

was asserted without justification.

The easiest indication that your argument doesn't go through is the fact that you never used compactness.

Here's a direct proof . . .

Since $\text{diam}\;K_n\ge x$, and each $K_n$ is compact, we can choose $p_n,q_n\in K_n$ such that $d(p_n,q_n) \ge x$.

Since $K_1\supseteq K_n$ for all $n$, it follows that the sequences $(p_n)$ and $(q_n)$ are sequences in $K_1$.

Hence, since $K_1$ is compact, each of the sequences $(p_n),(q_n)$ has at least one limit point in $K_1$.

Let $u,v\in K_1$ be limit points of $(p_n),(q_n)$, respectively.

By continuity of the distance function, we have $d(u,v)\ge x$.

If we can show $u,v\in K$, we're done.

Fix a positive integer $m$.

Since $u$ is a limit point of the sequence $(p_n)$, it follows that $u$ is also a limit point of the subsequence 
$$p_m,p_{m+1},p_{m+2},...$$
all terms of which are in $K_m$.

Hence, since $K_m$ is compact, it follows that $u\in K_m$.

Since the positive integer $m$ was arbitrary, it follows that $u\in K_m$ for all positive integers $m$, hence $u\in K$.

By analogous reasoning, we get $v\in K$.

Therefore $\text{diam}\;K\ge x$, as was to be shown.
