To construct a set with a limit point.

Construct a set with exactly one limit point and then add two distinct copies of it; let me be more clear: you'll surely agree with the fact that $\{0\} \cup \{{\frac{1}{n} : n \in \mathbb{N} ^{*}}\}$ has only one limit point. Thus if we repeat twice a translation we'll get what we're looking for: $(\{0\} \cup \{{\frac{1}{n} : n \in \mathbb{N} ^{*}}\}) \sqcup (\{2\} \cup \{{\frac{1}{n}+2 : n \in \mathbb{N} ^{*}}\}) \sqcup (\{4\} \cup \{{\frac{1}{n} +4: n \in \mathbb{N} ^{*}}\})$.

Hint: What are the limit points of $\left\{\frac{1}{n}\middle|\ n\in \mathbb{N}\right\}$?

Take a sequence $\{ a_n \}_{n \in \mathbb{N}}$ converging to $0$, another sequence $\{ b_n \}_{n \in \mathbb{N}}$ converging to $\frac{1}{2}$, and another sequence $\{ c_n \}_{n \in \mathbb{N}}$ converging to $1$. Then just take the set $$ S := \{ a_n,b_n,c_n; n \in \mathbb{N} \} $$ to be their union. By definition $0,\frac{1}{2}$ and $1$ will be limit points. For example, you can let $a_n := \frac{1}{n}$, $b_n := \frac{1}{2} - \frac{1}{2^n}$, $c_n := 1 - \frac{1}{n!}$. This choice also conforms to your wish that $S \subseteq [0,1]$.