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Yes. For every non-isolated $p\in X$, there is a well-ordered net in $X\setminus\{p\}$ converging to $p$ as long as $X$ is compact and Hausdorff. This result was observed in the 1992 paper Convergent free sequences in compact spaces by I. Juhász and Z. Szentmiklóssy, so let me paraphrase their argument. Suppose that $\lambda$ is a limit ordinal and $(A_\alpha)_{\alpha<\lambda}$ is a descending sequence of closed subsets of a compact Hausdorff space $X$ with $\bigcap_{\alpha<\lambda}A_\alpha=\{p\}$ but where $|A_\alpha|>1$ for $\alpha<\lambda$. Let $x_\alpha\in A_\alpha\setminus\{p\}$ for $\alpha<\lambda$. Then whenever $U$ is a neighborhood of $p$, by compactness, there is some $\alpha<\lambda$ where $x_\beta\in A_\beta\subseteq U$ whenever $\lambda>\beta\geq\alpha$. Therefore, $x_\alpha\rightarrow p$.

We still need to construct the sequence $(A_\alpha)_{\alpha<\lambda}$. Let $\lambda$ be the smallest cardinal such that there exists a collection of closed sets $(C_\alpha)_{\alpha<\lambda}$ where $\{p\}=\bigcap_{\alpha<\lambda}C_\alpha$ but where $p\in (C_\alpha)^\circ$ for each $\alpha<\lambda$. We can therefore set $A_\alpha=\bigcap_{\beta<\alpha}C_\beta$.

We also observe that some form of compactness is necessary for this argument to work; if $p\in\beta\mathbb{N}\setminus\mathbb{N}$, then there is no well-ordered net in $\mathbb{N}$ that converges to $p$.

Unique accumulation makes the proof work

Compact Hausdorff spaces can be characterized as the Hausdorff spaces where each net that accumulates at a unique point $p$ actually converges to the point $p$. This characterization of compact Hausdorff spaces allows us to break our result up into a couple of lemmas.

We say that a filter $\mathcal{F}$ on a topological space $X$ accumulates at the point $p\in X$ if $p\in\bigcap_{R\in\mathcal{F}}\overline{R}$. We say that a net $(x_{d})_{d\in D}$ on the same space $X$ accumulates at the point $p\in X$ the filter generated by $\{\{x_{e}\mid e\geq d\}\mid d\in D\}$ accumulates at the point $p$. Equivalently, the net $(x_d)_{d\in D}$ accumulates at $p$ precisely when $U\cap\{x_e\mid e\geq d\}\neq\emptyset$ whenever $U$ is a neighborhood of $p$.

Lemma: Let $X$ be a Hausdorff space. Then the following are equivalent:

1. $X$ is compact.

2. If $(x_d)_{d\in D}$ is a net that converges at the unique point $p\in X$, then $(x_d)_{d\in D}$ converges to the point $p.$

Lemma: Let $X$ be a topological space. Let $p\in X$ be a non-isolated point. Suppose that $\{p\}=\bigcap_{U\in\mathcal{N}(p)}\overline{U}$ where $\mathcal{N}(p)$ is the filter of neighborhoods of $p$ (this is a weakening of regularity). Then there is a cardinal $\lambda$ and well-ordered net $(x_\alpha)_{\alpha<\lambda}$ where if $(x_\alpha)_{\alpha<\lambda}$ accumulates at a point $q$, then $q=p$.

Proof: Let $\lambda$ be the smallest cardinal such that there exists a transfinite sequence $(U_\alpha)_{\alpha<\lambda}$ of neighborhoods of $p$ with $\bigcap_{\alpha<\lambda}\overline{U_\alpha}=\{p\}$. Then since $\lambda$ is minimized, the set $\bigcap_{\beta\leq\alpha}\overline{U_\beta}\setminus\{p\}$ is non-empty for each $\alpha<\lambda$. Then set $x_\alpha\in\bigcap_{\beta\leq\alpha}\overline{U_\beta}\setminus\{p\}$ for $\alpha<\lambda$. Therefore, $\bigcap_{\beta<\lambda}\overline{\{x_\alpha\mid \alpha\geq\beta\}}\subseteq\bigcap_{\beta<\lambda}\overline{U_\beta}=\{p\}.$ $\square$

By combining the two above lemmas, we can conclude that for each point $p$ in a compact Hausdorff space $X$, there is a well-ordered net in $X\setminus\{p\}$ converging to $p$.

Yes. For every non-isolated $p\in X$, there is a well-ordered net in $X\setminus\{p\}$ converging to $p$ as long as $X$ is compact and Hausdorff. This result was observed in the 1992 paper Convergent free sequences in compact spaces by I. Juhász and Z. Szentmiklóssy, so let me paraphrase their argument. Suppose that $\lambda$ is a limit ordinal and $(A_\alpha)_{\alpha<\lambda}$ is a descending sequence of closed subsets of a compact Hausdorff space $X$ with $\bigcap_{\alpha<\lambda}A_\alpha=\{p\}$ but where $|A_\alpha|>1$ for $\alpha<\lambda$. Let $x_\alpha\in A_\alpha\setminus\{p\}$ for $\alpha<\lambda$. Then whenever $U$ is a neighborhood of $p$, by compactness, there is some $\alpha<\lambda$ where $x_\beta\in A_\beta\subseteq U$ whenever $\lambda>\beta\geq\alpha$. Therefore, $x_\alpha\rightarrow p$.

We still need to construct the sequence $(A_\alpha)_{\alpha<\lambda}$. Let $\lambda$ be the smallest cardinal such that there exists a collection of closed sets $(C_\alpha)_{\alpha<\lambda}$ where $\{p\}=\bigcap_{\alpha<\lambda}C_\alpha$ but where $p\in (C_\alpha)^\circ$ for each $\alpha<\lambda$. We can therefore set $A_\alpha=\bigcap_{\beta<\alpha}C_\beta$.

We also observe that some form of compactness is necessary for this argument to work; if $p\in\beta\mathbb{N}\setminus\mathbb{N}$, then there is no well-ordered net in $\mathbb{N}$ that converges to $p$.

Unique accumulation makes the proof work

Compact Hausdorff spaces can be characterized as the Hausdorff spaces where each net that accumulates at a unique point $p$ actually converges to the point $p$. This characterization of compact Hausdorff spaces allows us to break our result up into a couple of lemmas.

We say that a filter $\mathcal{F}$ on a topological space $X$ accumulates at the point $p\in X$ if $p\in\bigcap_{R\in\mathcal{F}}\overline{R}$. We say that a net $(x_{d})_{d\in D}$ on the same space $X$ accumulates at the point $p\in X$ the filter generated by $\{\{x_{e}\mid e\geq d\}\mid d\in D\}$ accumulates at the point $p$. Equivalently, the net $(x_d)_{d\in D}$ accumulates at $p$ precisely when $U\cap\{x_e\mid e\geq d\}\neq\emptyset$ whenever $U$ is a neighborhood of $p$.

Lemma: Let $X$ be a Hausdorff space. Then the following are equivalent:

1. $X$ is compact.

2. If $(x_d)_{d\in D}$ is a net that accumulates at the unique point $p\in X$, then $(x_d)_{d\in D}$ converges to the point $p.$

Lemma: Let $X$ be a topological space. Let $p\in X$ be a non-isolated point. Suppose that $\{p\}=\bigcap_{U\in\mathcal{N}(p)}\overline{U}$ where $\mathcal{N}(p)$ is the filter of neighborhoods of $p$ (this is a weakening of regularity). Then there is a cardinal $\lambda$ and well-ordered net $(x_\alpha)_{\alpha<\lambda}$ where if $(x_\alpha)_{\alpha<\lambda}$ accumulates at a point $q$, then $q=p$.

Proof: Let $\lambda$ be the smallest cardinal such that there exists a transfinite sequence $(U_\alpha)_{\alpha<\lambda}$ of neighborhoods of $p$ with $\bigcap_{\alpha<\lambda}\overline{U_\alpha}=\{p\}$. Then since $\lambda$ is minimized, the set $\bigcap_{\beta\leq\alpha}\overline{U_\beta}\setminus\{p\}$ is non-empty for each $\alpha<\lambda$. Then set $x_\alpha\in\bigcap_{\beta\leq\alpha}\overline{U_\beta}\setminus\{p\}$ for $\alpha<\lambda$. Therefore, $\bigcap_{\beta<\lambda}\overline{\{x_\alpha\mid \alpha\geq\beta\}}\subseteq\bigcap_{\beta<\lambda}\overline{U_\beta}=\{p\}.$ $\square$

By combining the two above lemmas, we can conclude that for each point $p$ in a compact Hausdorff space $X$, there is a well-ordered net in $X\setminus\{p\}$ converging to $p$.

Yes. This result was observed in the 1992 paper Convergent free sequences in compact spaces by I. Juhász and Z. Szentmiklóssy, so let me paraphrase their argument. Suppose that $\lambda$ is a limit ordinal, and $(A_\alpha)_{\alpha<\lambda}$ is a descending sequence of closed subsets of a compact Hausdorff space $X$ with $\bigcap_{\alpha<\lambda}A_\alpha=\{p\}$. Let $x_\alpha\in A_\alpha\setminus A_{\alpha+1}$ for $\alpha<\lambda$. Then whenever $U$ is a neighborhood of $p$, by compactness, there is some $\alpha<\lambda$ where $x_\beta\in A_\beta\subseteq U$ whenever $\lambda>\beta\geq\alpha$. Therefore, $x_\alpha\rightarrow p$.

Yes. For every non-isolated $p\in X$, there is a well-ordered net in $X\setminus\{p\}$ converging to $p$ as long as $X$ is compact and Hausdorff. This result was observed in the 1992 paper Convergent free sequences in compact spaces by I. Juhász and Z. Szentmiklóssy, so let me paraphrase their argument. Suppose that $\lambda$ is a limit ordinal and $(A_\alpha)_{\alpha<\lambda}$ is a descending sequence of closed subsets of a compact Hausdorff space $X$ with $\bigcap_{\alpha<\lambda}A_\alpha=\{p\}$ but where $|A_\alpha|>1$ for $\alpha<\lambda$. Let $x_\alpha\in A_\alpha\setminus\{p\}$ for $\alpha<\lambda$. Then whenever $U$ is a neighborhood of $p$, by compactness, there is some $\alpha<\lambda$ where $x_\beta\in A_\beta\subseteq U$ whenever $\lambda>\beta\geq\alpha$. Therefore, $x_\alpha\rightarrow p$.

We still need to construct the sequence $(A_\alpha)_{\alpha<\lambda}$. Let $\lambda$ be the smallest cardinal such that there exists a collection of closed sets $(C_\alpha)_{\alpha<\lambda}$ where $\{p\}=\bigcap_{\alpha<\lambda}C_\alpha$ but where $p\in (C_\alpha)^\circ$ for each $\alpha<\lambda$. We can therefore set $A_\alpha=\bigcap_{\beta<\alpha}C_\beta$.

We also observe that some form of compactness is necessary for this argument to work; if $p\in\beta\mathbb{N}\setminus\mathbb{N}$, then there is no well-ordered net in $\mathbb{N}$ that converges to $p$.

Unique accumulation makes the proof work

Compact Hausdorff spaces can be characterized as the Hausdorff spaces where each net that accumulates at a unique point $p$ actually converges to the point $p$. This characterization of compact Hausdorff spaces allows us to break our result up into a couple of lemmas.

We say that a filter $\mathcal{F}$ on a topological space $X$ accumulates at the point $p\in X$ if $p\in\bigcap_{R\in\mathcal{F}}\overline{R}$. We say that a net $(x_{d})_{d\in D}$ on the same space $X$ accumulates at the point $p\in X$ the filter generated by $\{\{x_{e}\mid e\geq d\}\mid d\in D\}$ accumulates at the point $p$. Equivalently, the net $(x_d)_{d\in D}$ accumulates at $p$ precisely when $U\cap\{x_e\mid e\geq d\}\neq\emptyset$ whenever $U$ is a neighborhood of $p$.

Lemma: Let $X$ be a Hausdorff space. Then the following are equivalent:

1. $X$ is compact.

2. If $(x_d)_{d\in D}$ is a net that converges at the unique point $p\in X$, then $(x_d)_{d\in D}$ converges to the point $p.$

Lemma: Let $X$ be a topological space. Let $p\in X$ be a non-isolated point. Suppose that $\{p\}=\bigcap_{U\in\mathcal{N}(p)}\overline{U}$ where $\mathcal{N}(p)$ is the filter of neighborhoods of $p$ (this is a weakening of regularity). Then there is a cardinal $\lambda$ and well-ordered net $(x_\alpha)_{\alpha<\lambda}$ where if $(x_\alpha)_{\alpha<\lambda}$ accumulates at a point $q$, then $q=p$.

Proof: Let $\lambda$ be the smallest cardinal such that there exists a transfinite sequence $(U_\alpha)_{\alpha<\lambda}$ of neighborhoods of $p$ with $\bigcap_{\alpha<\lambda}\overline{U_\alpha}=\{p\}$. Then since $\lambda$ is minimized, the set $\bigcap_{\beta\leq\alpha}\overline{U_\beta}\setminus\{p\}$ is non-empty for each $\alpha<\lambda$. Then set $x_\alpha\in\bigcap_{\beta\leq\alpha}\overline{U_\beta}\setminus\{p\}$ for $\alpha<\lambda$. Therefore, $\bigcap_{\beta<\lambda}\overline{\{x_\alpha\mid \alpha\geq\beta\}}\subseteq\bigcap_{\beta<\lambda}\overline{U_\beta}=\{p\}.$ $\square$

By combining the two above lemmas, we can conclude that for each point $p$ in a compact Hausdorff space $X$, there is a well-ordered net in $X\setminus\{p\}$ converging to $p$.

Yes. We can always find a convergent linearly ordered net. Let $X$ be a compact Hausdorff space, and let $p\in X$. Then let $\lambda$ be the smallest cardinal such that there exists open neighborhoods $(V_\alpha)$ of $p$ with $\{p\}=\bigcap_{\alpha<\lambda}V_\alpha$. Furthermore, assume that $U_\alpha$ is an open set with $p\in U_\alpha\subseteq\overline{U_\alpha}\subseteq V_\alpha$ for $\alpha<\lambda$. Then let $x_\alpha\in(\bigcap_{\beta<\alpha}U_\alpha)\setminus\{p\}$ for $\alpha<\lambda$. I claim that $(x_\alpha)_{\alpha<\lambda}\rightarrow p$. Let $U$ be an open neighborhood of $p$. Then $\{p\}=\bigcap_{\alpha<\lambda}\overline{U_\alpha}\subseteq U$, so $X=U\cup(\bigcap_{\alpha<\lambda}\overline{U_\alpha})^c=U\cup\bigcup_{\alpha<\lambda}(\overline{U_\alpha})^c$. Therefore, by compactness, there are $\alpha_1,\dots,\alpha_r$ with $X=U\cup(\overline{U_{\alpha_1}})^c\cup\dots\cup(\overline{U_{\alpha_r}})^c$. In this case, $\overline{U_{\alpha_1}}\cap\dots\cap\overline{U_{\alpha_r}}\subseteq U$. However, if $\alpha>\max(\alpha_1,\dots,\alpha_r)$, then $x_\alpha\in U$. We may now conclude that $(x_\alpha)_{\alpha<\lambda}\rightarrow p$.

Yes. This result was observed in the 1992 paper Convergent free sequences in compact spaces by I. Juhász and Z. Szentmiklóssy, so let me paraphrase their argument. Suppose that $\lambda$ is a limit ordinal, and $(A_\alpha)_{\alpha<\lambda}$ is a descending sequence of closed subsets of a compact Hausdorff space $X$ with $\bigcap_{\alpha<\lambda}A_\alpha=\{p\}$. Let $x_\alpha\in A_\alpha\setminus A_{\alpha+1}$ for $\alpha<\lambda$. Then whenever $U$ is a neighborhood of $p$, by compactness, there is some $\alpha<\lambda$ where $x_\beta\in A_\beta\subseteq U$ whenever $\lambda>\beta\geq\alpha$. Therefore, $x_\alpha\rightarrow p$.