A theorem of Demailly and Pǎun (https://arxiv.org/abs/math/0105176) gives a condition for a compact complex manifold to be of Fujiki class $\mathcal{C}$ (that is bimeromoprhic to a Kähler manifold). A manifold is Kähler if and only if it supports a Kähler current. A Kähler current is a real positive $(1,1)$-current $T$, such that $T - \varepsilon \omega > 0$ for a constant $\varepsilon > 0$ and positive hermitian $(1,1)$-form $\omega$.
The homology class of such current is integral if and only if the manifold is projective (this is the standard Kodaira embedding theorem).
A similar theorem for Moishezon manifolds was proved by Popovici (https://arxiv.org/abs/math/0603738). It states, roughly speaking, that a complex manifold is Moishezon if and only if it supports a semi-positive and strictly positive almost everywhere real $(1,1)$-current with integral homology class.
The integrality condition is hardly a differential-geometric one, but you cannot escape it (at least, as far, is we know by now), and I don't really think one might give a more differential-geometric answer.
In a general situation it is, of course, rather hard to verify the Popovici condition. Another "geometric" approach to studying algebraic dimensions of complex manifolds is to study the algebraic reduction. For any complex manifold $X$ there exists a normal projective variety $\overline{X}$ and a meromorphic map $\alpha \colon X \to \overline{X}$, such that any meromorphic function on $X$ can be lifted from $\overline{X}$. The variety $\overline{X}$ is unique up to birational equivalence.
Being Moishezon is equivalent to $\alpha$ being a birational equivalence. More generally, $a(X) = \dim_{\mathbb C}(\overline{X})$.
As mentioned in the comments, now I think the argument in the book is incorrect, at least with the given assumption. The following was my first attempt to answer this question, where the crossed sentences are insufficient:
Every compact Kahler manifold $X$ admits a Kahler form $[\omega]\in H^2(X,\mathbb C)$ which is nonzero. Moreover, $[\omega]$ is real and of type $(1,1)$. Therefore, by the assumption that $\dim_{\mathbb C}H^{1,1}(X)=1$, $[\omega]$ is a real element of the complex vector space $\mathbb C\cong H^{1,1}(X)$. One can rescale $[\omega]$ by a constant $r\in\mathbb R^+$, such that $r[\omega]\in H^2(X,\mathbb Z)$ is integral and still stay in the Kahler cone.
For your confusion, you can check from the fact $H^{1,1}(X)$ is a complex subspace of $H^{2}(X,\mathbb C)$ such that $\overline{H^{1,1}(X)}=H^{1,1}(X)$, it follows that $$H^{1,1}(X)= \big (H^{1,1}(X)\bigcap H^2(X,\mathbb R)\big )\otimes_{\mathbb R} \mathbb C.$$
So in particular, the intersection $\require{enclose}\enclose{horizontalstrike}{H^2(X,\mathbb Z)\cap H^{1,1}(X)\neq \{0\}}$.
The reason of insufficiency is the following: Since $H^{1,1}(X)$ is 1-dimensional, the Kahler cone is a ray $\mathbb R^+$ in the real vector space $H^2(X,\mathbb R)$, so $X$ is a Hodge manifold iff the that ray passes an (nonzero) integral element in $H^2(X,\mathbb R)$. Explicitly, let's say $h^{2,0}=h^{0,2}=1$, and $v_1,v_2,v_3$ is a basis of $H^2(X,\mathbb Z)$, then $[\omega]=\sum_{i=1}^3r_iv_i$ for some $r_i\in \mathbb R$, can we gurantee that $[r_1:r_2:r_3]=[n_1:n_2:n_3]$ with $n_i$ integer? No, and general weight two Hodge structure won't satisfy that.
Best Answer
Yes, this boils down to two facts, which you should be able to find in e.g. Huybrecht's Complex Geometry or Voisin's Hodge Theory and Complex Algebraic Geometry: I.
$\mathbb{P}^n(\mathbb{C})$ is a Kähler manifold. This is true, since we can endow it with the Fubini-Study metric.
If $M$ is a Kähler manifold and $N\subset M$ is a complex submanifold, then $(N,\omega|_N)$ is Kähler, where $\omega$ is the Kähler form of $M$.
Now every smooth projective variety $X$ over $\mathbb{C}$ admits a closed embedding $X\hookrightarrow \mathbb{P}^n$ (as schemes over $\mathbb{C}$). Applying the analytification functor realizes $X(\mathbb{C})$ as a complex submanifold of $\mathbb{P}^n(\mathbb{C})$ and combining the two above facts you are done.