Let $X$ be a compact Kähler manifold, $\mathcal K_X$ be the Kähler cone of $X$, if we have $H^{1,1}(X) \cap H^2(X,\mathbb Z)\neq 0$, then can we get $\mathcal K_X \cap H^2(X,\mathbb Z) \neq \emptyset$?
At first, I think it is true because I have a "proof": since the Kähler cone $\mathcal K_X$ forms an open set in $H^{1,1}(X,\mathbb R)$ (see p130 of Huybrechts'《complex geometry》), and $H^{1,1}(X) \cap H^2(X,\mathbb Z)\neq 0$ implies $H^{1,1}(X,\mathbb R) \cap H^2(X,\mathbb Q)\neq 0$, we know rational points $\mathbb Q$ is dense in $\mathbb R$, so an open set $\mathcal K_X$ in $H^{1,1}(X,\mathbb R)$ also contains rational points in $H^2(X,\mathbb Q)$, thus $\mathcal K_X \cap H^2(X,\mathbb Q) \neq \emptyset$.
But as I have asked a related question here, I think I have gotten a counter-example provided by @Ariyan Javanpeykar: let $Y$ be a non-projective compact Kähler manifold, take $X=Y\times \mathbb CP^1$, then $X$ is not projective. And we know that $X$ is projective if and only if $\mathcal K_X \cap H^2(X,\mathbb Z) \neq \emptyset$ (cited from p251 of Huybrechts' 《complex geometry》), since $X$ is not projective, we get $\mathcal K_X \cap H^2(X,\mathbb Z) = \emptyset$. But compute the Neron-Severi group of $X$, we have $H^{1,1}(X) \cap H^2(X,\mathbb Z)\neq 0$, so it answers negatively my question.
Can anyone figure out what mistakes I have made in my "proof"? Thanks!