The $K$-topology on $\mathbb{R}$, generated by the open intervals together with their complements of $K = \{1/(n+1) : n \in \mathbb{N}\}$, is $T_1$ and Hausdorff but not regular

(B1): every $x \in \mathbb{R}$ lies in $(x-1, x+1) \in \mathcal{B}_K$, so $\mathcal{B}_K$ covers $\mathbb{R}$.

(B2): for $a < b$ and $c < d$ the intersection $(a,b) \cap (c,d)$ is $(\max\{a,c\}, \min\{b,d\})$ by [L5] and [A1], which is a member of $\mathcal{B}_K$ when $\max\{a,c\} < \min\{b,d\}$ and is empty otherwise; removing $K$ from one or from both factors intersects the same interval with the complement of $K$, giving a member of $\mathcal{B}_K$ or the empty set; and in the empty case (B2) is vacuous.

By steps 1.1 and 1.2 and [L1] the family $\mathcal{B}_K$ is a basis for a unique topology $\mathcal{T}_K$ on $\mathbb{R}$, which is claim 1's first half.

$\mathcal{T}_K$ is finer than the usual topology: if $U$ is usually open and $x \in U$, then [L2] gives $r > 0$ with $x \in (x-r,x+r) \subseteq U$, and $(x-r,x+r) \in \mathcal{B}_K$, so $U \in \mathcal{T}_K$ by [L1]; this completes claim 1.

Let $x \ne y$ in $\mathbb{R}$; by [L5] we may assume $x < y$, and put $r := (y-x)/2 > 0$, so that $x + r = (x+y)/2 = y - r$. The sets $(x-r,x+r)$ and $(y-r,y+r)$ lie in $\mathcal{B}_K$, hence are open, contain $x$ and $y$ respectively, and are disjoint, a common point being both $< x+r$ and $> y-r = x+r$. So $(\mathbb{R},\mathcal{T}_K)$ is Hausdorff, and $T_1$ by [L6]; this is claim 2.

$\mathbb{R} \setminus K = \bigcup_{k \ge 1} \big( (-k, k) \setminus K \big)$, the union being over naturals $k \ge 1$: each term is a member of $\mathcal{B}_K$ and misses $K$, and conversely a point $x \notin K$ has $|x| < k$ for some natural $k \ge 1$ by [L3], so $x$ lies in the term of index $k$. Hence $\mathbb{R} \setminus K$ is open by [L7] and $K$ is closed, which is claim 3.

Suppose $U, V \in \mathcal{T}_K$ are disjoint with $0 \in U$ and $K \subseteq V$; note $0 \notin K$, since every element of $K$ is positive by [L4].

Under step 4.1: by [L1] there is $B \in \mathcal{B}_K$ with $0 \in B \subseteq U$, and $B$ is $(a,b)$ or $(a,b) \setminus K$ with $a < 0 < b$.

Under step 4.1: by [L3] there is a natural $k \ge 1$ with $1/k < b$, and $k = n+1$ for some $n \in \mathbb{N}$, so $1/(n+1) \in (a,b)$, since $0 < 1/(n+1) < b$ by [L4] and $a < 0$.

Under step 4.1: $B \ne (a,b)$, for otherwise $1/(n+1) \in B \subseteq U$ by step 6.1 while $1/(n+1) \in K \subseteq V$, contradicting $U \cap V = \varnothing$; hence $B = (a,b) \setminus K$.

Under step 4.1: $1/(n+1) \in K \subseteq V$, so by [L1] there is $B' \in \mathcal{B}_K$ with $1/(n+1) \in B' \subseteq V$; and $B'$ is not of the form $(c,d) \setminus K$, which contains no point of $K$, so $B' = (c,d)$ with $c < 1/(n+1) < d$.

Under step 4.1: put $t := \max\{\, \max\{a,c\},\ 1/(n+2) \,\}$ by [L5]. Then $t < 1/(n+1)$: indeed $a < 0 < 1/(n+1)$ and $c < 1/(n+1)$, and $1/(n+2) < 1/(n+1)$ by [L4], since $\iota(n+1) < \iota(n+2)$ and both are positive.

Under step 4.1: the interval $(t,\ 1/(n+1))$ contains no element of $K$. An element $1/(m+1)$ of it satisfies $1/(m+1) < 1/(n+1)$, hence $\iota(n+1) < \iota(m+1)$ by [L4], hence $n + 1 < m + 1$ and so $m + 1 \ge n + 2$, whence $1/(m+1) \le 1/(n+2) \le t$ by [L4] and step 8.1, contradicting $t < 1/(m+1)$.

Under step 4.1: put $z := (t + 1/(n+1))/2$, so $t < z < 1/(n+1)$ by [A1] and step 8.1, and $z \notin K$ by step 9.1.

Under step 4.1: $z \in (a,b) \setminus K = B \subseteq U$, since $a \le \max\{a,c\} \le t < z$ and $z < 1/(n+1) < b$ by step 6.1 and step 8.1; and $z \in (c,d) = B' \subseteq V$, since $c \le \max\{a,c\} \le t < z$ and $z < 1/(n+1) < d$ by step 7.2.

Step 11.1 puts $z$ in $U \cap V$, contradicting the disjointness assumed in step 4.1; so no such $U$ and $V$ exist, and by [L6] the space $(\mathbb{R},\mathcal{T}_K)$ is not regular, which is claim 4.