Context: Let $\Omega$ be an open set in $\mathbb C$ and $K$ be a compact subset of $\Omega$.
Question: find a $\alpha$ $\in$ $C^\infty_0(\Omega)$ such that it is 1 on $K$.
So far: I found that by Urysohn lemma we can find a continuous function with compact support and $\alpha$= 1 on K. And to get smoothness we can take convolution with standard mollifier. But I am not getting $1$ on $K$.
There are many ways to construct such functions. Your approach is excellent. Let us slightly tweak it to achieve the property your are missing.
First, for $r > 0$, let us define $$ K_r := \{ x \in \mathbb C \mid d(x,K) \leq r \}. $$ For each $r > 0$, $K_r$ is a also a compact set (which contains $K$). Moreover, you can prove (for example arguing by contradiction), that there exists $r_0 > 0$ such that $K_{r_0} \subset \Omega$.
Then, apply your initial argument to the compact $K_{r_0}$ to find a continuous function $f$ which is $1$ on $K_{r_0}$ and compactly supported in $\Omega$. Now, as you said, regularize it by convolution, letting for example $f_\epsilon := f \star \rho_\epsilon$ where $\rho_\epsilon(z) := \rho(z/\epsilon)/\epsilon^2$ and $\rho$ is a smooth mollifier.
Now if you start from a function $\rho$ which has compact support, (for example in the ball of radius 1), then you can check that, for $\epsilon < r_0$, $f_\epsilon = 1$ on $K$ (because for $x \in K$, $f_\epsilon(x)$ is computed as an average of values of $f$ over $K_{r_0}$).
And you get your conclusion, almost with your arguments.
For an explicit version of such a function, first define $$g(x):=\exp\frac{-1}{x(1-x)}\quad\text{for}\;0< x<1,$$ with $g(x):=0$ otherwise. Then $g\in\mathrm C^\infty(\Bbb R)$ (although $g\notin\mathrm C^\omega(\Bbb R)$). Let $$h(x):=\frac{\int_{-\infty}^xg(t)\,\mathrm dt}{\int_{-\infty}^\infty g(t)\,\mathrm dt}\quad(x\in\Bbb R).$$ Note that $h(x)=0$ for $x\leqslant0$, and $h(x)=1$ for $x\geqslant1$, with $h\in\mathrm C^\infty(\Bbb R)$. Now define $$f(x):=h(x)h(3-x)\quad(x\in\Bbb R).$$ Then $f\in\mathrm C^\infty(\Bbb R)$ and $f(x)=1$ for $1\leqslant x\leqslant2$, with $f(x)=0$ outside the interval $0< x<3$.
For a complex version of this, define $\alpha:\Bbb C\to\Bbb R:z\mapsto f(z\bar z),$ which satisfies $\alpha(z)=1$ inside the annulus $1\leqslant z\bar z\leqslant2$, with $\alpha(z)=0$ for $z\bar z\geqslant3$.
Added in edit: $\;$ For completeness, given an open $\varOmega\subset\Bbb C$, with compact $K\subset\varOmega$, there is a disk $D$, say with centre $c$, that includes $K$. Let $\beta:z\mapsto \surd\frac32+\epsilon (z-c)$ $\,(z\in \Bbb C$), where $\epsilon>0$ is small enough for $\beta$ to map all of $D$ into the annulus $\{z:1\leqslant z\bar z\leqslant2\}$. Then, for $\alpha$ as above, the map $\alpha':=\alpha\circ\beta$ has the required property as stated in the question. The function $\alpha'$ is zero outside the (compact) disk $D':=\{z: |z-(c-\surd\frac32/\epsilon)|\leqslant\surd3/\epsilon\}$.
