I am trying to prove that lower limit topology is completely normal. I know the it is normal. I attempted to consider this as cases let $X$ completely normal and $Y$ subset of $X$
If $Y$ countable then $Y$ has discrete topology.So, $Y$ is normal.
If $Y$ is closed, then $Y$ is normal.
But other cases, it is not obvious to me.
A very general result is true; I’ll leave just part of the proof to you. We’ll need the following result, which is proved here.
Let $\langle X,\le\rangle$ be any linear order, and let $\tau$ be the order topology on $X$. Then the space $\langle X,\tau\rangle$ is $T_5$ (i.e., completely normal and $T_1$).
Now let $\langle X,\le\rangle$ be a linear order, and let $\tau$ be the lower limit (or Sorgenfrey) topology on $X$. Let $Y=X\times\{0,1\}$, and let $\preceq$ be the lexicographic order on $Y$: $\langle x_0,i_0\rangle\preceq\langle x_1,i_1\rangle$ if and only if either $x_0<x_1$, or $x_0=x_1$ and $i_0<i_1$. Let $\tau_Y$ be the order topology on $Y$ induced by the order $\preceq$. Let $X_1=X\times\{1\}$, and let $\tau_1$ be the relative topology on $X_1$ as a subspace of $Y$.
Now we can apply the first result: $\langle Y,\tau_Y\rangle$ is $T_5$, so each of its subspaces is $T_5$, and in particular $\langle X_1,\tau_1\rangle$ and its homeomorphic copy $\langle X,\tau\rangle$ are $T_5$.
