Every infinite subset of $E$ in $\mathbb R^k$ having a limit point in $E$ implies $E$ is closed

Suppose $S$ is finite. Then the set $\{\, |\mathbf x - \mathbf x_o| : \mathbf x \in S \,\}$ is a finite set of strictly positive numbers. Thus it has a smallest member, which is a positive number. Choose $m$ so large that $1/m$ is less than that smallest number. Then no points of $E$ are within a distance $1/m$ of $\mathbf x_o,$ so $\mathbf x_o$ would not be a limit point.

Let's parse the part about having "constant positive value for infinitely many $n.$"

If a set is finite, then a sequence of members of that set has "constant positive value for infinitely many" terms. For example, consider the digits $0,1,2,3,4,5,6,7,8,9.$ The decimal expansion of an irrational number cannot contain only finitely many of each of those.

If a particular positive number occurs infinitely many times in a sequence, then the sequence cannot approach $0.$ But the sequence of distances $( |\mathbf x_n - \mathbf x_o| )_{n=1}^\infty$ approaches $0.$

That last inequality holds if $n$ is large enough, since in that case $1/n$ is small.

Suppose $S$ is finite, say $S=\{y_1,\cdots,y_k\}\subseteq\{x_n:n \in \mathbb N\}\subseteq E.$ Then the possible values each $|x_n-x_0|$ can take are $$|y_1-x_0|,\cdots,|y_k-x_0|.$$

So there will exists $i \in \{1,\cdots,k\}$ such that $|x_n-x_0|=|y_i-x_0|$ for infinitely many $n.$ So we have $$|y_i-x_0|\leq \frac 1n$$ for infinitely many $n.$ Thus, $x_0 = y_i \in E,$ which is a contradiction.

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Since, $x_0 \neq y,$ then there exists $n_0 \in \mathbb N$ such that $$|x_0-y|>\frac {2}{n_0}>\frac 2n$$ for all $n \geq n_0.$ Thus, $$\frac 12 |x_0 -y|>\frac 1n$$ for all $n \geq n_0.$

So, $$|x_0-y|-\frac 12 |x_0-y|>\frac 1n$$ for all $n\geq n_0.$ Hence, the last inequality follows.

If $S$ was finite, then you can take minimum of $|x_n - x_0|$, since there are finitely many $x_n\in S$. (minimum is not $0$ since $x_0\notin S$) But this contradicts with the fact that there are $x_n$'s satisfying $|x_n-x_0| < 1/n$ for all $n$.

For the last inequality;

\begin{equation} |\mathbf{x}_o-\mathbf{y}| - \dfrac{1}{n} \geq \dfrac{1}{2}|\mathbf{x}_o-\mathbf{y}| \iff \dfrac{1}{2}|\mathbf{x}_o-\mathbf{y}| \geq \dfrac{1}{n} \end{equation} and this is true for all but finitely many $n$, since $\dfrac{1}{2}|\mathbf{x}_o-\mathbf{y}|$ is a fixed positive number and $1/n\to 0$.