The minus symbol in the expression $(a,b)-K$ denotes relative complement, not element-wise subtraction.
Thus
\begin{align*}
(-1,1)-K
&=(-1,1)\setminus K\\[4pt]
&=
(-1,0]
\cup\left(
\bigl({\small{\frac{1}{2}}},1\bigr)
\cup
\bigl({\small{\frac{1}{3}}},{\small{\frac{1}{2}}}\bigr)
\cup
\bigl({\small{\frac{1}{4}}},{\small{\frac{1}{3}}}\bigr)
\cup
\cdots
\right)
\end{align*}
By definition, any open interval is open in the $K$-topology, hence any open set in the standard topology is still open in the $K$-topology.
To see that the $K$-topology is strictly finer than the standard topology, simply note that no open interval containing the element $0$ is a subset of $(1,1)-K$, so $(1,1)-K$ is not open in the standard topology.
If $\tau'$ is strictly finer than $\tau$, then the subspace topology $\sigma'$ induced on $Y \subseteq X$ by $\tau'$ will be finer than the one, $\sigma$, induced by $\tau$. Why? Well, any open set $V$ in $\sigma$ is of the form $U \cap Y$ where $U$ is in $\tau$. But, $\tau'$ is strictly finer than $\tau$ i.e. $\tau \subset \tau'$. Hence $U$ is in $\tau'$ also; so we managed to write $V$ as the intersection of an open set in $\tau'$ with $Y$ i.e. $U$ belongs to $\sigma'$. Thus, $\sigma \subseteq \sigma'$ and not $\sigma \subset \sigma'$. So something definitely went wrong in your example above using $\mathbb{R}_{st}$ and $\mathbb{R}_\ell$ and $\mathscr{B}_1$ and $\mathscr{B}_2$. Let us discuss that:
Therefore, every element of $\mathscr{B}_{1}$ is contained in $\mathscr{B}_{2}$.
Strictly speaking, that is not true. But you don't need that. What is true is if $B \in \mathscr{B}_1$ contains $x$, then there is an element $B' \in \mathscr{B}_2$ such that $x \in B' \subseteq B$. This is exactly what condition (ii)(b) requires. For example, if $B = (b', 1]$, and $x \in (b', 1]$, then you can just take $B' = [x, 1]$ to get $x \in [x, 1] \subseteq (b', 1]$. Next, if $B = (a'', b'') \ni x$, then you can take $B' = [x, b'')$ to get $x \in [x, b'') \subseteq (a'', b'')$ again. So on and so forth. And this leads to:
Thus, using (ii), I conclude that $\sigma ' \subset \sigma$.
You are using (ii) in exactly the opposite manner. Given what I explained above about $\mathscr{B}_1$ and $\mathscr{B}_2$, you actually get according to (ii)(a) that $\sigma'$ is finer than $\sigma$, which is just another way of saying $\sigma \subseteq \sigma'$ as one should expect.
Extra Fact:
In general $\sigma \subseteq \sigma'$ will not be strict. You can indeed have $\sigma = \sigma'$ even though $\tau \subset \tau'$ strictly. The topology you considered, $\mathbb{R}_\ell$, is not necessarily the best to demonstrate this. Instead, consider the $K$-topology $\mathbb{R}_K$ on $\mathbb{R}$ which is also mentioned in Munkres. It has as basis all open intervals $(a, b)$ just like $\mathbb{R}_{st}$. But its basis also contains sets of the form $(a, b) - K$ where $K = \{\frac{1}{n} \in \mathbb{R} \ : n \in \mathbb{Z}_+\}$.
Note what is going on. The set $K$ only affects open intervals that overlap with $[0, 1]$. So informally the topology of $\mathbb{R}_K$ away from that region is exactly the same as $\mathbb{R}_{st}$. So if for example $Y = [6, 10]$ or any other subset that does not overlap with $[0, 1]$, its subspace topology under $\mathbb{R}_K$ will be the same as the subspace topology induced by the standard topology. For if you have a basis element of $\sigma'$ of the form $((a, b) - K) \cap Y$, then because $Y$ has no elements in common with $K$, we have $((a, b) - K) \cap Y = (a, b) \cap Y$. So basis elements of $\sigma'$ coincide with those of $\sigma$.
Best Answer
Of $O\in\mathcal{T}$, then $O$ can be written as the union of intervals of the type $(a,b)$. Each interval $(a,b)$, in turn, can be written as$$\bigcup_{x\in(a,b)}[x,b),$$which belongs to $\mathcal{T}_1$. Therefore, $O\in\mathcal{T}_1$.