Let $\varphi\in{\mathcal{C}}^{\infty}(X,\mathbb{R})$ be such that $(1-\varepsilon)\omega\,|_{{}_{X}}+dd^{c}\varphi\geq 0$ on $X$, for some $\varepsilon>0$. We first choose $\tilde{\varphi}$ to be any smooth extension of $\varphi$ to $V$. Consider

$$\psi:=\tilde{\varphi}+A\chi\,\text{dist}(\cdot,X)^{2},$$

where $\chi$ is a test function supported in a small neighborhood of $X$ and such that $\chi\equiv 1$ near $X$. Here ${dist}$ is any Riemannian distance on $V$, for instance the distance associated to the Kähler metric $\omega$. Then $\psi$ is yet another smooth extension of $\varphi$ to $V$, which now satisfies $(1-\varepsilon/2)\omega+dd^{c}\psi\geq 0$ near $X$, if $A$ is chosen large enough.

The function $\log(\text{dist}(\cdot,X)^{2})$ is well defined and qpsh in a neighborhood of $X$. Let $\chi$ be a test function supported in this neighborhood so that $\chi\equiv 1$ near $X$. The function $u=\chi\log(\text{dist}(\cdot,X)^{2})$ is $N\omega$-psh on $V$ for a large integer $N$. Moreover, $\exp(u)$ is smooth and $X=\{u=-\infty\}$. Replacing $\omega$ by $N\omega$, $\varphi$ by $N\varphi$, and $\psi$ by $N\psi$, we may assume that $N=1$. Set now

$$\psi_{C}:=\frac{1}{2}\,\log\left[e^{2\psi}+e^{u+C}\right].$$

This again is a smooth extension of $\varphi$, and a straightforward computation yields

$$dd^{c}\psi_{C}\geq\frac{2e^{2\psi}dd^{c}\psi+e^{u+C}dd^{c}u}{2(e^{2\psi}+e^{u+C})}\;.$$

Hence

$$\left(1-\frac{\varepsilon}{2}\right)\omega+dd^{c}\psi_{C}\geq\frac{2e^{2\psi}\left[\left(1-\frac{\varepsilon}{2}\right)\omega+dd^{c}\psi\right]+(1-\varepsilon)e^{u+C}\omega}{2(e^{2\psi}+e^{u+C})}\geq 0,$$

if $C$ is chosen large enough. ∎