In Voinsin's book [1], Theorem 11.32 (page 280) says: 


"If X is an algebraic variety, these subgroups of $Hdg^{2k}(X) coincide." 


However, the proof did not show that the subgroup generated by
cycle classes (denoted by $A$) is containded in the subgroup generated by Chern classes of vector bundles (denoted by $B$) when $k>1$.
In fact it just claims that $B\subseteq A$.


There are two questions:


(1) How to show $A\subseteq B$?


(2) Why is the condition that $X$ is an algebraic variety necessary? 


[1] C. Voisin, Hodge theory and complex algebraic geometry, Vol. I, Cambrige Univ Press, 2002



Why do the subgroups of $Hdg^{2k}(X)$ generated by the cycle classes and Chern classes of vector bundles  coincide in algebraic variety $X$?
To expand slightly on Minhyong's comment, the key facts can be
found in Fulton's Intersection Theory. If you look at the comment following
corollary 18.3.2, you'll see an isomorphism (in slightly different notation)
$$ch:K^0(X)\otimes \mathbb{Q}\cong CH(X)\otimes\mathbb{Q}$$
where $X$ is a nonsingular variety, $K^0(X)$ is the Grothendieck group of
vector bundles, $CH(X)$ is the Chow group of cycles mod rational equivalence,
and $ch$ is the Chern character. After mapping this to rational cohomology, you get exactly
the statement you want.


Note that this is false if


1. you omit the $\mathbb{Q}$, or
2. if you work on a general (compact) complex manifold because there may not be enough vector bundles or subvarieties. To be clear, I mean that conclusion in cohomology is false:
In Zucker, "Hodge conjecture for cubic fourfolds" Compositio 1977, you can find an example 
of a torus with a nonzero integral $(1,1)$ class which is not a divisor, but it would
necessarily lie in the image of $c\_1$.


