(C'') $\iff$ (C''') by taking complements.

(C) $\Rightarrow$ (C'): Assume (C).
To show: If $𝒢$ is an ultrafilter on $X$ then there exists $x\in X$ such that $x$ is a limit point of $𝒢\text{.}$
Assume $𝒢$ is an ultrafilter on $X\text{.}$
By (C), there exists $x\in X$ such that $x$ is a cluster point of $𝒢\text{.}$
Since $𝒢$ is an ultrafilter $x$ is a limit point of $𝒢\text{.}$

(C') $\Rightarrow$ (C): Assume (C').
To show: If $𝒥$ is a filter on $X$ then there exists $x\in X$ such that $x$ is a cluster point of $𝒥\text{.}$
Assume $𝒥$ is a filter on $X\text{.}$
Since the collection of filters on $X$ satisfies the hypotheses of Zorn's lemma, there exists an ultrafilter $𝒢$ such that $𝒢\supseteq 𝒥\text{.}$
By (C') there exists $x\in X$ such that $x$ is a limit point of $𝒢\text{.}$
So $x$ is a cluster point of $𝒢\text{.}$
Since $𝒢\supseteq 𝒥$ and $x$ is a cluster point of $𝒢$ then $x$ is a cluster point of $𝒥\text{.}$

(not C'') $\Rightarrow$ (not C): Assume that there is a collection $𝒞$ of closed sets such that $\bigcap _{K\in 𝒞}K=\varnothing$ and, if $\ell \in {\mathbb{Z}}_{>0}$ and $K_1,\dots ,K_{\ell }\in 𝒞$ then $K_1\cap \cdots \cap K_{\ell }\ne \varnothing \text{.}$ Let $𝒥$ be the set of subsets of $X$ which contain a set in $𝒞\text{.}$
Then $𝒥$ is a filter.
Since $\bigcap _{N\in 𝒥}\bar{N}\subseteq \bigcap _{K\in 𝒞}\bar{K}=\bigcap _{K\in 𝒞}K=\varnothing ,$ $𝒥$ does not have a cluster point.

(not C) $\Rightarrow$ (not C''): Assume that there exists a filter $𝒥$ on $X$ with no cluster point.
Then $\bigcap _{N\in 𝒥}\bar{N}=\varnothing \text{.}$
Since $𝒥$ is a filter, if $\ell \in {\mathbb{Z}}_{>0}$ and $N_1,\dots ,N_{\ell }\in 𝒥$ then $N_1\cap \cdots \cap N_{\ell }\ne \varnothing$ and therefore $\overline{N_1}\cap \cdots \cap \overline{N_{\ell }}\ne \varnothing \text{.}$
Let $𝒞=\left\{\bar{N}\hspace{0.17em}\right|\hspace{0.17em}N\in 𝒥\}\text{.}$
Then $𝒞$ is a collection of closed sets such that $\bigcap _{K\in 𝒞}K=\varnothing$ but there does not exist $K_1,\dots ,K_{\ell }\in 𝒞$ such that $K_1\cap K_2\cap \cdots \cap K_{\ell }=\varnothing \text{.}$

$\square$

Let $X$ be a Hausdorff topological space with topology $𝒯\text{.}$ Let $K\subseteq X\text{.}$
Assume $K$ is compact.
To show: $K$ is closed.
To show: $K^c$ is open.
To show: If $x\in K^c$ then $x$ is an interior point of $K^c\text{.}$
Assume $x\in K^c\text{.}$
To show: There exists $U\in 𝒯$ such that $x\in U$ and $U\subseteq K^c\text{.}$
For each $y\in K$ there exist $U_{xy}\in 𝒯$ and $V_{xy}\in 𝒯$ such that $x\in U_{xy}$ and $y\in V_{xy}$ and $U_{xy}\cap V_{xy}=\varnothing \text{.}$ Then $$\left\{V_{xy}\hspace{0.17em}\right|\hspace{0.17em}y\in K\}$$ is an open cover of $X\text{.}$
Since $K$ is compact there exists $N\in {\mathbb{Z}}_{>0}$ and $y_1,y_2,\dots ,y_N\in K$ such that $$\{V_{x{y}_1},V_{x{y}_2},\dots ,V_{x{y}_N}\}$$ is an open cover of $K\text{.}$
Let $$U=U_{x{y}_1}\cap \cdots \cap U_{x{y}_N}\text{.}$$ Since $U$ is a finite intersection of open sets, $U$ is open. Also $x\in U$ and $$\begin{array}{ccc}{\displaystyle U\cap K}& =& {\displaystyle (U_{x{y}_1}\cap \cdots \cap U_{x{y}_N})\cap (V_{x{y}_1}\cup \cdots \cup V_{x{y}_N})}\\ & =& {\displaystyle \varnothing \text{.}}\end{array}$$ So $x\in U$ and $U\subseteq K^c\text{.}$
So $x$ is an interior point of $K^c\text{.}$
So $K^c$ is open.
So $K$ is closed.

1. Let $x\in \bar{K}$. The neighbourhood filter $ℬ\left(x\right)$ of $x$ induces a filter ${ℬ}_K$ WHAT DOES THIS MEAN??? on $K$ which has a cluster point $y\in ????$. Since $ℬ\left(x\right)$ is coarser than ${ℬ}_K$ (considered as a filter base on $X$) the point $y$ is a cluster point of $ℬ\left(x\right)$. So $y=x$ since $X$ is Hausdorff.

$\square$