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, $\operatorname{diam}K \geq x.$
Best Answer
No, your proof is flawed.
Your last line
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.