Where does my confusion arise from? I am getting $H^{(1,1)}(X,\mathbb{Z})=NS(X)$, the Neron-Severi group, however in any reference I don't find this remarkable thing.
Let $X$ be a Kahler manifold. Using the exponential sequence one obtains a homomorphism $H^1(X,\mathcal{O}_X^*)\rightarrow H^2(X,\mathbb{Z})$. This is associating to a holomorphic line bundle its Chern class $c_1$.
If we take any connection $D$ on a holomorphic line bundle $L$, if we do $D\circ D$ and then take its class in $H^2(X,\mathbb{C})$ we get a curvature form independent of the choice of the connection. Multiplying by $-i/(2\pi)$ we run into $c_1(L)$ when identifying $H^2(X,\mathbb{Z})$ as a subspace of $H^2(X,\mathbb{C})$. This is a theorem. The last in particualr shows that the image of $c_1$ is always $(1,1)$ form.
On the other hand, the map $H^1(X, O_X^*)\rightarrow H^{(1,1)}(X,\mathbb{Z})$ is surjective by the Lefschetz theorem.
So $\operatorname{im} c_1 =H^{(1,1)}(X,\mathbb{Z})$ ? The group of Neron-Severi $NS:=H^1(X,O_X^*)/\ker c_1\cong \mathrm{im} c_1$
Of course, you should be writing $H^{(1,1)}(X)\cap H^2(X,\Bbb Z)$ when you use your shorthand. And you need $X$ compact Kähler. But ... You're almost right: The Lefschetz Theorem assumes $X$ is a projective variety, does it not?