$\mathcal{T}_1:$ That's right, well done. Note that this topology is generated by the intervals $(a,b)$. So for any point $x$ in $\mathbb{R}$ in between $k = \frac1n$ and $k^\prime = \frac1m$ we can choose $(x - m, x + m)$ where $m = \frac12 \min (\frac1n , \frac1m)$ to see that $x$ is not in the closure of $K$. Looking at $0$ we see that every interval $0 \in (a,b)$ has to have $0 < b$ so there will be an $n$ such that $0 < \frac1n < b$ and hence every open set containing $0$ will have non-empty intersection with $K$. Hence we get $\overline{K} = K \cup \{0\}$.
$\mathcal{T}_2$: Note that $\mathcal{T}_2$ is generated by $(a,b) $ and $ (a,b) \setminus K $ so $\mathcal{T}_1 \subset \mathcal{T}_2$ and hence $ \overline{K}_{\mathcal{T}_2} \subset \overline{K}_{\mathcal{T}_1}$ since we have more sets that might contain our point but might not intersect with $K$. Since $K \subset \overline{K}$ we therefore know that the only point we need to think about is $0$. But none of the sets we added to the basis intersect with $K$ hence there now are open sets $O$ containing $0$ with $O \cap K = \varnothing$. Hence $0$ is not in the closure of $K$ and we have $\overline{K}=K$.
$\mathcal{T}_3:$ Note that the open sets in this topology look like this: $O = \mathbb{R} \setminus \{x_1, \dots, x_n \}$. Consider a point $p$ in $\mathbb{R}$. Let $O$ be an open set containing $p$. Note that $O$ has non-empty intersection with $K$ since $O$ contains all points of $\mathbb{R}$ except finitely many. In particular, $K$ contains infinitely many points hence their intersection will be non-empty. Since $p$ was arbitrary we get $\overline{K} = \mathbb{R}$.
$\mathcal{T}_4$: Note that the upper limit topology is generated by intervals of the form $(a, b]$. Note that for a point $p \neq 0$ outside $K$ we can pick $(\varepsilon , p]$ with $\varepsilon$ small enough such that $(\varepsilon , p] \cap K = \varnothing$. For $0$ we pick $(a, 0]$ to get a set that does not intersect with $K$ hence $0$ is not in the closure of $K$. Hence we get $\overline{K} = K$.
$\mathcal{T}_5$: Again you have it right, well done. To see this pick a point $p$ to the right of $0$ or $0$ itself and note that every open interval $(-\infty, a)$ containing $p$ will intersect with $K$ since there will be an $n$ such that $\frac1n < p$.
Hope this helps.
Let $\tau$ be the lexicographic order topology on $\Bbb R^2$ and $\tau_s$ the lexicographic order topology on the unit square $S=[0,1]\times[0,1]$. The main thing that you’re missing, I think, is that $\tau_s$ is not the topology that $S$ inherits as a subspace of $\langle\Bbb R^2,\tau\rangle$.
In $\langle\Bbb R^2,\tau\rangle$ each vertical $\{x\}\times\Bbb R$ is a clopen set, and in fact the lexicographically ordered plane is homeomorphic to the product $\Bbb R_d\times\Bbb R$, where $\Bbb R_d$ denotes $\Bbb R$ with the discrete topology, and the second factor $\Bbb R$ has the usual topology. Since $\Bbb R_d$ is metrizable with the discrete metric, this is a product of two metrizable spaces and as such is metrizable. One compatible metric is given by
$$d(\langle x_0,y_0\rangle,\langle x_1,y_1\rangle)=\begin{cases}
\min\{|y_0-y_1|,1\},&\text{if }x_0=x_1\\
1,&\text{otherwise}\;.
\end{cases}$$
In the relative topology induced on $S$ by $\tau$ the sets $\{x\}\times[0,1]$ for $x\in[0,1]$ would be clopen, and the restriction of $d$ to $S\times S$ would be a metric inducing the relative topology.
In $\tau_s$, however, those sets are not clopen: they’re closed, but they’re not open. If $0<a\le 1$, every open $\tau_s$-nbhd of $\langle a,0\rangle$ contains points $\langle x,0\rangle$ with $x<a$, and if $0\le a<1$, every $\tau_s$-open nbhd of $\langle x,1\rangle$ contains points $\langle x,1\rangle$ with $x>a$. Thus, none of the sets $\{x\}\times[0,1]$ with $x\in[0,1]$ is open in $\tau_s$. It’s this difference between $\tau_s$ and the relative topology inherited from $\tau$ that makes $S$ compact in the topology $\tau_s$, as it clearly is not in the relative topology, in which
$$\big\{\{x\}\times[0,1]:x\in[0,1]\big\}$$
is an open cover of $S$ with no finite subcover — indeed, no proper subcover.
In short, you can start with a compatible metric on the $\langle\Bbb R^2,\tau\rangle$ and restrict it to a metric on $S$, and it will generate the subspace topology on $S$, but this isn’t $\tau_s$, the lexicographic order topology on $S$. Once you realize that you’re looking at two different topologies on $S$, the apparent problem disappears.
As an aside, it’s worth noting that the two topologies actually do agree at all points of $S$ except those of the form $\langle x,0\rangle$ with $0<x\le 1$ and those of the form $\langle x,1\rangle$ with $0\le x<1$.
Best Answer
Let $L$ be the image of the map $\beta: (-\pi,\pi) \to \Bbb R^2$, a "lemniscate", where $\beta(t)=(\sin 2t, \sin t)$, see this question for a picture. Define the map $f: \Bbb R \to L$ by $\beta(2\arctan(x))$ and note that this is continuous (reals in the usual topology), and 1-1.
Also let $Y = [-\infty,+\infty]{/}\{-\infty, +\infty,0\}$ in the quotient topology induced by the identification map $q: [-\infty,+\infty]$ that identifies the three points, and the extended reals have their standard topology with basic neighbourhoods $[-\infty, n), n \in \Bbb Z$ for the left end point and $(n,+\infty]$ for the right end point. We can extend $f$ to $\hat{f}: [-\infty,+\infty] \to L$ by defining $\hat{f}(\pm \infty) = 0$ as well, which is continuous and then $Y \simeq L$ as $q$ factorises through it.
There remains a simple verification that $L \simeq \Bbb R$ where the latter has this looped line topology. $f$ essentially becomes the homeomorphism.