Answer
See explanation
Work Step by Step
A standard way to see that **every infinite set \(S\) contains a countably infinite subset** is to construct an infinite sequence of distinct elements of \(S\). Concretely:
1. **Pick an element** \(s_1 \in S\).
2. **Remove** \(s_1\) from \(S\), leaving \(S_1 = S \setminus \{s_1\}\). Because \(S\) is infinite, \(S_1\) is still nonempty.
3. **Pick another element** \(s_2 \in S_1\) with \(s_2 \neq s_1\).
4. **Remove** \(s_2\) from \(S_1\), leaving \(S_2 = S_1 \setminus \{s_2\}\).
5. **Continue** in this fashion, always picking a new element \(s_n\) from \(S_{n-1}\).
Since \(S\) is infinite, you can keep choosing new elements indefinitely. This process produces a sequence
\[
s_1,\; s_2,\; s_3,\;\dots
\]
of **distinct** elements of \(S\). The set \(\{s_1, s_2, s_3, \dots\}\) is then **countably infinite** and is a **subset** of \(S\).
Hence every infinite set has (at least) one countably infinite subset.
