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This $X$ also has a subbase of clopen sets, and is not sequential because it has a non-sequential closed subspace, the Arens-Fort space $(\omega\times\omega\times\{\omega\})\cup\{*\}.$ As before, $X\setminus\{*\}$ is sequential. To show that $(\omega\times\omega\times\omega)\cup\{*\}$ is sequential, consider a set $C\subseteq \omega\times\omega\times\omega$ such that $*$ is a limit point of $C.$ Define $C_\gamma=\{(x,y,z)\in C\mid y\geq f_\alpha(x)\}.$

This $X$ also has a subbase of clopen sets, and is not sequential because it has a non-sequential closed subspace, the Arens-Fort space $(\omega\times\omega\times\{\omega\})\cup\{*\}.$ As before, $X\setminus\{*\}$ is sequential. To show that $(\omega\times\omega\times\omega)\cup\{*\}$ is sequential, consider a set $C\subseteq \omega\times\omega\times\omega$ such that $*$ is a limit point of $C.$ Define $C_\gamma=\{(x,y,z)\in C\mid y\geq f_\gamma(x)\}.$

I believe there is a regular non-sequential almost sequential space in ZFC+CH.

Now assume that $*$ is not a limit point of $C_\gamma,$ for some $\gamma.$ Take $\gamma$ to be minimal. So $*$ is a limit point of $C_{\beta}$ if and only if $\beta<\gamma.$ Since $0\neq\gamma<\omega_1$ there is a countable sequence $\beta_n$ with $\sup(\beta_n+1)=\gamma.$ For each $n,N$ the set $C_{\beta_n}\setminus C_\gamma$ must intersect the neighborhood $U_{\gamma,N},$ which means there are $(x,y)\in C_{\beta_n}\setminus C_\gamma$ with $x>N$ and $y\leq f_\gamma(x).$ Therefore $C_{\beta_1}\cap\dots\cap C_{\beta_n}\setminus C_\gamma$ (which might be smaller than $C_{\beta_n}\setminus C_\gamma$ at a finite number of $x$-coordinates) contains some $(x_n,y_n)$ with $x>N$ and $y\leq f_\gamma(x).$ We can pick such a choice of $(x_n,y_n)$ for each $n,$ using $N$ to ensure that $x_n$ is strictly increasing. This construction ensures that for each $\beta<\gamma$ the sequence $(x_n,y_n)$ eventually lies in $C_{\beta},$ and $y_n\leq f_\gamma(x_n)$ for all $n.$ So for each $\beta,n$ the sequence $(x_n,y_n)$ eventually lies in $U_{\beta,n}.$ This proves that $(x_n,y_n)$ converges to $*.$

I believe there is a regular non-sequential almost sequential space in ZFC+CH. (See below for a construction using a weaker assumption.)

Now assume that $*$ is not a limit point of $C_\gamma,$ for some $\gamma.$ Take $\gamma$ to be minimal. So $*$ is a limit point of $C_{\beta}$ if and only if $\beta<\gamma.$ Since $0\neq\gamma<\omega_1$ there is a countable sequence $\beta_n$ with $\sup(\beta_n+1)=\gamma.$ For each $n,N$ the set $C_{\beta_n}\setminus C_\gamma$ must intersect the neighborhood $U_{\gamma,N},$ which means there are $(x,y)\in C_{\beta_n}\setminus C_\gamma$ with $x>N$ and $y\leq f_\gamma(x).$ Therefore $C_{\beta_1}\cap\dots\cap C_{\beta_n}\setminus C_\gamma$ (which might be smaller than $C_{\beta_n}\setminus C_\gamma$ at a finite number of $x$-coordinates) contains some $(x_n,y_n)$ with $x_n>N$ and $y_n\leq f_\gamma(x_n).$ We can pick such a choice of $(x_n,y_n)$ for each $n,$ using $N$ to ensure that $x_n$ is strictly increasing. This construction ensures that for each $\beta<\gamma$ the sequence $(x_n,y_n)$ eventually lies in $C_{\beta},$ and $y_n\leq f_\gamma(x_n)$ for all $n.$ So for each $\beta,N$ the sequence $(x_n,y_n)$ eventually lies in $U_{\beta,N}.$ This proves that $(x_n,y_n)$ converges to $*.$

If I understand correctly, the above argument relied on $\mathfrak{u}=\mathfrak{d}=\aleph_1.$ I believe this assumption can be weakened to $\mathfrak{d}=\aleph_1,$ which is used for the diagonalization at the end.

The argument is very similar. We still have $f_\alpha$ but no ultrafilter nor $S_\alpha.$ I will assume $f_0(n)=0.$ Pick a bijective function $p:\omega\times\omega\to\omega.$ The base set for $X$ will instead be $(\omega\times\omega\times(\omega+1))\cup\{*\},$ and $U_{\alpha,N}$ will instead be $$U_{\alpha,N}=\{*\}\cup\{(x,y,z)\mid x>N\text{ and either }y\geq f_\alpha(x) \text{ or }z\leq f_\alpha(p(x,y))\}.$$

This $X$ also has a subbase of clopen sets, and is not sequential because it has a non-sequential closed subspace, the Arens-Fort space $(\omega\times\omega\times\{\omega\})\cup\{*\}.$ As before, $X\setminus\{*\}$ is sequential. To show that $(\omega\times\omega\times\omega)\cup\{*\}$ is sequential, consider a set $C\subseteq \omega\times\omega\times\omega$ such that $*$ is a limit point of $C.$ Define $C_\gamma=\{(x,y,z)\in C\mid y\geq f_\alpha(x)\}.$

First consider the case that $*$ is a limit point of $C_\gamma$ for every $\gamma.$ Let $D=\{(x,y)\mid \exists z.(x,y,z)\in C\}.$ Pick any function $f:\omega\to\omega$ with $(x,y,f(p(x,y)))\in C$ for each $(x,y)\in D.$ Pick $\alpha$ such that $f\leq^* f_\alpha.$ Pick a sequence of pairs $(x_n,y_n)\in D$ with $x_n\to\infty$ and $y_n\geq f_\alpha(x_n)$ (these exist because $C_\alpha$ is not bounded in the $x$ direction). Consider an arbitrary $U_{\beta,N}.$ We either have $\beta\leq\alpha$ giving $y_n\geq f_\beta(x_n)$ eventually, or $\beta\geq\alpha$ giving $f(p(x_n,y_n))\leq f_\beta(p(x_n,y_n))$ eventually. This proves that $(x_n,y_n,f(p(x_n,y_n)))$ converges to $*.$

Now assume that $*$ is not a limit point of $C_\gamma,$ for some $\gamma.$ Take $\gamma$ to be minimal. So $*$ is a limit point of $C_{\beta}$ if and only if $\beta<\gamma.$ Since $0\neq\gamma<\omega_1$ there is a countable sequence $\beta_n$ with $\sup(\beta_n+1)=\gamma.$ For each $n,N$ the set $C_{\beta_n}\setminus C_\gamma$ must intersect the neighborhood $U_{\gamma,N},$ which means there are $(x,y,z)\in C_{\beta_n}\setminus C_\gamma$ with $x>N$ and $z\leq f_\gamma(p(x,y)).$ Therefore $C_{\beta_1}\cap\dots\cap C_{\beta_n}\setminus C_\gamma$ (which might be smaller than $C_{\beta_n}\setminus C_\gamma$ at a finite number of $x$-coordinates) contains some $(x_n,y_n,z_n)$ with $x_n>N$ and $z_n\leq f_\gamma(p(x_n,y_n)).$ We can pick such a choice of $(x_n,y_n,z_n)$ for each $n,$ using $N$ to ensure that $x_n$ is strictly increasing. This construction ensures that for each $\beta<\gamma$ the sequence $(x_n,y_n,z_n)$ eventually lies in $C_{\beta},$ and $z_n\leq f_\gamma(p(x_n,y_n))$ for all $n.$ So for each $\beta,N$ the sequence $(x_n,y_n,z_n)$ eventually lies in $U_{\beta,N}.$ This proves that $(x_n,y_n,z_n)$ converges to $*.$

The construction will make use of an ultrafilter $\mathcal U$ on $\omega,$ a cofinal increasing $\omega_1$-sequence $S_\alpha$ in $(\mathcal U,\supseteq^*),$ and a cofinal increasing $\omega_1$ sequence $f_\alpha$ in $(\omega^\omega,\leq^*).$ So $\alpha<\beta$ implies $S_\alpha\supset^* S_\beta,$ and for every set $S$ there is $\alpha$ such that $S\supseteq^* S_\alpha$ or $\omega\setminus S\supseteq^* S_\alpha.$ And $\alpha<\beta$ also implies $f_\alpha\leq^*f_\beta,$ and for every $f$ there is $\alpha$ such that $f\leq^* f_\alpha.$ I will also require $S_0=\omega.$ These are easy to construct under CH by transfinite induction. Specifically, we can take $S_\alpha$ to be a strictly $\subseteq^*$-decreasing subsequence of the sets called $X_\alpha$ in the construction of a Ramsey ultrafilter in Jech's Set theory Theorem 7.8 (3rd ed), and take $f_\alpha(n)$ to be smallest integer with $|S_\alpha\cap\{0,1,\dots,f_\alpha(n)\}|=n.$

The construction will make use of an ultrafilter $\mathcal U$ on $\omega,$ a cofinal increasing $\omega_1$-sequence $S_\alpha$ in $(\mathcal U,\supseteq^*),$ and a cofinal increasing $\omega_1$ sequence $f_\alpha$ in $(\omega^\omega,\leq^*).$ So $\alpha<\beta$ implies $S_\alpha\supset^* S_\beta,$ and for every set $S$ there is $\alpha$ such that $S\supseteq^* S_\alpha$ or $\omega\setminus S\supseteq^* S_\alpha.$ And $\alpha<\beta$ also implies $f_\alpha\leq^*f_\beta,$ and for every $f$ there is $\alpha$ such that $f\leq^* f_\alpha.$ I will also require $S_0=\omega.$ These are easy to construct under CH by transfinite induction. (Specifically, we can take $S_\alpha$ to be a strictly $\subseteq^*$-decreasing subsequence of the sets called $X_\alpha$ in the construction of a Ramsey ultrafilter in Jech's Set theory Theorem 7.8 (3rd ed). $f_\alpha(n)$ can be constructed by a very similar argument.)