Definition 1: A space $X$ has a strong rank 1-diagonal \cite{5} if there exists a sequence $\{\mathcal U_n: n\in \omega\}$ of open covers of $X$ such that for each $x\in X$, $\{x\}=\bigcap \{\overline{\operatorname{St}(x, \mathcal U_n}): n\in \omega\}$.
Example 2: Let $Y=\bigcup\{[0,1]\times\{n\}:n \in \omega\}$ and $X=Y\cup\{a\}$ where $a \notin Y.$ Define a basis for a topology on $X$ as follows. Basic open sets containing $a$ take the form $\{a \}\cup \bigcup \{[0,1)\times {m}: m \geq n \}$ where $n \in \omega.$ Basic open sets about the other points of $X$ are the usual induced metric open sets.
Question 3: Is $X$ has a strong rank 1-diagonal?
Proof: Construct a sequence $\{\mathcal V_n: n\in \omega\}$ of open covers of $Y$ which witnesses that $Y$ has a strong rank 1-diagonal. Let $\mathcal U_n= \mathcal V_n \cup (\{a \}\cup \bigcup \{[0,1)\times {m}: m \geq n \})$. Then $\{\mathcal U_n: n\in \omega\}$ shows that $X$ has a strong rank 1-diagonal.
I'm not sure the proof is OKey. Could somebody help me?
Your argument is fine. You could even replace $[0,1)$ in $\{a\}\cup\bigcup\big\{[0,1)\times\{m\}:m\ge n\big\}$ with $[0,1]$, as can be seen directly or by the following argument.
$X$ is obtained in an especially simple fashion from a compact metric space, which of course has a strong rank $1$ diagonal. Let me construct it a little differently. For $n\in\omega$ let $$Y_n=\left[0,\frac1{2^n}\right]\times\left\{\frac1{2^n}\right\}\;,$$ let $$Y=\bigcup_{n\in\omega}Y_n\;,$$ let $a=\langle 0,0\rangle$, and let $X=Y\cup\{a\}$. Let $\tau$ be the topology that $X$ inherits from the plane; $\langle X,\tau\rangle$ is evidently a compact metric space.
Now let $D=\left\{\left\langle 0,\frac1{2^n}\right\rangle:n\in\omega\right\}$, and let $\tau'$ be the topology generated by the subbase $\tau\cup\{X\setminus D\}$; it’s not hard to check that $\tau'=\tau\cup\{U\setminus D:U\in\tau\}$, and that $\langle X,\tau'\rangle$ is homeomorphic to your space $X$. It is of course not regular, since $a$ cannot be separated from the closed set $D$.
However, it has the same closures of open sets as $\langle X,\tau\rangle$: if $U\in\tau$, then $\operatorname{cl}_\tau U=\operatorname{cl}_{\tau'}U$ and $\operatorname{cl}_\tau (U\setminus D)=\operatorname{cl}_{\tau'}(U\setminus D)$. It follows immediately that a family of $\tau$-open covers witnessing the fact that $\langle X,\tau\rangle$ has a strong rank $1$ diagonal does the same for $\langle X,\tau'\rangle$.