[step:Build the Levi-flat hypersurface and the CR function]
Write points of $\mathbb C^n$ as $z=(z',z_n)$, where $z'=(z_1,\dots,z_{n-1})\in\mathbb C^{n-1}$ and $z_n=s+it$ with $s,t\in\mathbb R$. Define
\begin{align*}
M:=\{(z',z_n)\in\mathbb C^{n-1}\times\mathbb C:\operatorname{Im}z_n=0\}.
\end{align*}
Let
\begin{align*}
p:=(0,\dots,0)\in M.
\end{align*}
The function
\begin{align*}
\rho:\mathbb C^n&\to\mathbb R
\end{align*}
\begin{align*}
(z',z_n)&\mapsto \operatorname{Im}z_n
\end{align*}
is a $C^\infty$ defining function for $M$, and $d\rho\ne0$ on $M$. Hence $M$ is a $C^\infty$ embedded real hypersurface. Its complex tangent bundle is spanned by the coordinate vector fields $\partial_{\bar z_1},\dots,\partial_{\bar z_{n-1}}$, and therefore its Levi form is identically zero. Thus $M$ is Levi-flat.
Define
\begin{align*}
u:M&\to\mathbb C
\end{align*}
\begin{align*}
(z',s)&\mapsto a(s).
\end{align*}
Since $a\in C^\infty((-1,1);\mathbb C)$, after multiplying by a smooth cutoff in the $s$ variable away from $0$ if desired, $u$ is a $C^\infty$ function on a neighbourhood of $p$ in $M$, and it extends to a global smooth function on $M$. For each $j\in\{1,\dots,n-1\}$,
\begin{align*}
\partial_{\bar z_j}u=0
\end{align*}
on $M$, because $u$ depends only on $s=\operatorname{Re}z_n$. Hence $u$ is CR on $M$.
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