In general you have the right ideas, and good on you for showing proper work! Some remarks:
You do not need to consider the topologies 1 and 4 separately, this will already fall out ot considering 1 and 2 and 2 and 4. If we have strict inclusion there, then a fortiori we will have it for 1 and 4, no need for a separate proof. Similary with 2 and 5, e.g.
What about 4 and 5? You just say "clear" but some argument seems in order.
You seem to assume that $x_1 < x_2$, because the open interval $(x_1,x_2)$ is non-empty (in 3 vs 5)? A bit sloppy.
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
You made a mistake in the following statement.
It should be $x \in B\subset U$.
Then use (iii) to get the desired result.
For proving ${\mathscr{T}}_{2}\nsubseteq {\mathscr{T}}_{1}.$
For any $B=(-\infty,a)\in {\mathscr{T}}_{2}$ there does not exist an open set $U \in{\mathscr{T}}_{1}$ such that $U\subset B$.
On contrary, we assume that there exists an open set $U \in{\mathscr{T}}_{1}$ such that $U\subset B$ holds.
$\implies [a,\infty)=\mathbb{R} \setminus B.$
$\implies [a,\infty) \subset \mathbb{R} \setminus U$ which is not finite.
It is a contradiction to the fact that $U \in{\mathscr{T}}_{1}$.
$\implies {\mathscr{T}}_{2}\nsubseteq {\mathscr{T}}_{1}$
Hence ${\mathscr{T}}_{2}$ and ${\mathscr{T}}_{1}$ are non-comparable.
Thank You @IEm for helpful comments.