Compact Sets Have Finite Diameter -- Trouble with Proof

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I'm working through Kaplansky's Set Theory and Metric Spaces. On page 101, Theorem 68 reads

"A compact metric space $M$ has finite diameter. The diameter is attained for two suitable points of $M$."

Here's the proof:

"Let t be the diameter of $M$. For the moment we allow $t=\infty$. We can find points $x_i, y_i \in M$ such that $D(x_i, y_i)$ approaches $t$. Since $M$ is compact, we can drop down to convergent subsequeces; so (after changing notation) we may assume that $x_i \rightarrow x$, $y_i \rightarrow y$. Then (Theorem 37) $D(x_i, y_i) \rightarrow D(x,y)$, so $t$ is finite and equals $D(x,y)$."

My problem lies with this notion of 'dropping down'. Sure, compactness gives us convergent subsequences, but why are these subsequences guaranteed to have distance approaching $t$? I don't see the impossibility of the subsequence $D(x_i, y_i)$ approaching something finite -- namely $D(x,y)$ -- while the diameter $t$ remains infinite. What am I missing?

Thank you.

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Do it in two stages. First find an infinite $N_0\subseteq\Bbb N$ such that $\langle x_k:k\in N_0\rangle$ converges to some $x\in M$. Then find an infinite $N_1\subseteq N_0$ such that $\langle y_k:k\in N_1\rangle$ converges to some $y\in M$; the subsequence $\langle x_k:k\in N_1\rangle$ of $\langle x_k:k\in N_0\rangle$ still converges to $x$, and the subsequence $\langle D(x_k,y_k):k\in N_1\rangle$ still converges to $t$. It also converges to $D(x,y)$, so $D(x,y)=t$.

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The key thing is that the distance function $D : M \times M \to [0,\infty)$ is continuous, so if $(x_i,y_i) \in M \times M$ approaches $(x,y)$ then $D(x_i,y_i)$ approaches $D(x,y)$.

But knowing that, there is another, more direct proof. The product $M \times M$ is compact (by the Tychonoff theorem), and the distance function $D : M \times M \to [0,\infty)$ is continuous (as just said), therefore $D$ has a maximum value (by the extreme value theorem). That maximum value is the diameter of $M$ (by definition of the diameter).