Here is Prob. 1, Sec. 28, in the book Topology by James R. Munkres, 2nd edition:
Give $[0, 1]^\omega$ the uniform topology. Find an infinite subset of this space that has no limit point.
Here is a Mathematics Stack Exchange post on this very problem.
My Attempt:
We note that $[0,1]^\omega$ denotes the set of all the sequences $\left( x_n \right)_{n \in \mathbb{N}}$ of the real numbers $x_n$ in the closed unit interval $[0,1]$, and that the uniform metric $\bar{\rho}$ on $[0,1]^\omega$ be given by $$ \bar{\rho}\left( \left( x_n \right)_{n\in\mathbb{N}} , \ \left( y_n \right)_{n\in\mathbb{N}} \right) \colon= \sup \left\{ \ \min \left\{ \ \left\vert x_n - y_n \right\vert, 1 \ \right\} \ \colon \ n \in \mathbb{N} \ \right\} \tag{A} $$ for all $\left(x_n\right)_{n \in \mathbb{N} }, \left(y_n \right)_{ n \in \mathbb{N} } \in [0,1]^\omega$.
We show that $[0,1]^\omega$ is not limit-point-compact
Let $$A \colon= \left\{ \ (x_n)_{n\in\mathbb{N}} \ \colon \ x_n \in \{ 0 , 1 \} \mbox{ for all } n \in \mathbb{N} \ \right\} = \{ 0, 1 \}^\omega. \tag{B} $$ This set $A$ is an infinite subset of $[0, 1]^\omega$.
We show that the set $A$ has no limit points in $[0,1]^\omega$.
Let $ \mathbf{x} \colon= (x_n)_{n\in\mathbb{N}}, \mathbf{y} \colon= (y_n)_{n\in\mathbb{N}} \in A$ be any two distinct points. Then $x_n, y_n \in \{ 0, 1\}$ for each $n \in \mathbb{N}$ and $x_k \neq y_k$ for at least one $k \in \mathbb{N}$, and for this $k$, we have $x_k - y_k = \pm 1$. Therefore we find that $$ \bar{\rho} \left( \mathbf{x}, \mathbf{y} \right) \geq \left\lvert x_k - y_k \right\rvert = 1, $$ but we know from (A) above that $$ \bar{\rho} \left( \mathbf{x}, \mathbf{y} \right) \leq 1; $$ therefore $$ \bar{\rho} ( \mathbf{x}, \mathbf{y} ) = \bar{\rho}\left( (x_n)_{n\in\mathbb{N}}, (y_n)_{n\in\mathbb{N}} \right) = 1. \tag{1} $$ for any two distinct points $\mathbf{x}, \mathbf{y} \in A$.
Let $\mathbf{p} \colon= \left( p_n \right)_{n \in \mathbb{N} }$ be any point of $[0, 1]^\omega$. We now show that this point $\mathbf{p}$ cannot be a limit point of set $A$.
Case 1. Suppose that $\mathbf{p} \in A$. Let us take any real number $\epsilon$ such that $0 < \epsilon < 1$. Then for any point $\mathbf{a} \in A$ such that $\mathbf{a} \neq \mathbf{p}$, we must have
$$ \bar{\rho} ( \mathbf{a}, \mathbf{p} ) = 1 > \epsilon, $$ because of (1) above. Therefore $$ \left( A \cap B_{\bar{\rho}} ( \mathbf{p}, \epsilon ) \right) \setminus \{ \mathbf{p} \} = \emptyset. $$ So $\mathbf{p}$ is not a limit point of $A$.Case 2. Suppose that $\mathbf{p} \in [0, 1]^\omega \setminus A$. Then there exists at least one $N \in \mathbb{N}$ for which $p_N \not\in \{ 0, 1\}$; for that $N$ we have $0 < p_N < 1$. Now let us take any real number $\epsilon$ such that $$ 0 < \epsilon < \min \left\{ \ \frac{p_N}{2}, \frac{1-p_N}{2} \ \right\}. $$ Then $$ \epsilon < \frac{p_N}{2} \qquad \mbox{ and } \qquad \epsilon < \frac{1-p_N}{2}, $$ and so $$ - \frac{p_N}{2} < - \epsilon \qquad \mbox{ and } \qquad \epsilon < \frac{1-p_N}{2}, $$ which in turn implies that $$ \frac{p_N}{2} = p_N - \frac{p_N}{2} < p_N - \epsilon \qquad \mbox{ and } \qquad p_N + \epsilon < p_N + \frac{1 - p_N}{2} = \frac{p_N + 1}{2}. $$ In fact we obtain $$ 0 < \frac{p_N}{2} < p_N - \epsilon \qquad \mbox{ and } \qquad p_N + \epsilon < \frac{p_N + 1}{2} < 1. \tag{2}$$ Now if $\mathbf{x} \colon= \left( x_n \right)_{n \in \mathbb{N} }$ in $[0, 1]^\omega$ satisfies $$ \bar{\rho}( \mathbf{x}, \mathbf{p} ) < \epsilon, $$ then for that point $\mathbf{x}$ we also have $$ \left\lvert x_N - p_N \right\rvert < \epsilon, $$ which implies that $$ p_N - \epsilon < x_N < p_N + \epsilon, $$ which together with (2) above implies that $$ 0 < \frac{p_N}{2} < p_N - \epsilon < x_N < p_N + \epsilon < \frac{p_N + 1}{2} < 1, $$ and thus $$ 0 < x_N < 1, $$ from which it follows that $\mathbf{x} \not\in A$. Thus we can conclude that $$ A \cap B_{\bar{\rho}}( \mathbf{p}, \epsilon ) = \emptyset, \tag{3}$$ and so $$ \left( A \cap B_{\bar{\rho}}( \mathbf{p}, \epsilon )\right) \setminus \{ \mathbf{p} \} = \emptyset. $$ Thus $\mathbf{p}$ is again not a limit point of our set $A$. In fact, $\mathbf{p}$ is not even an adherent point of set $A$.
From (3) we can even show that our set $A$ is in fact a closed set in $[0, 1]^\omega$.
Am I right?
Finally, it is my understanding that the above proof can go through even if we take our set $A$ as follows: Let $\alpha$ and $\beta$ be any two distinct real numbers such that $\alpha \in [0, 1]$ and $\beta \in [0, 1]$. Then let us put $$ A \colon= \left\{ \ \left( x_n \right)_{n \in \mathbb{N}} \ \colon \ x_n \in \{ \alpha, \beta \} \mbox{ for all } n \in \mathbb{N} \ \right\} = \{ \alpha, \beta \}^\omega. \tag{B*} $$
Is my proof correct and clear in all its details? If not, then where is it in need of improvement?