"König's theorem" for $T_2$-spaces? For any topological space $(X,\tau)$ we define a matching to be a collection of non-empty and pairwise disjoint open sets. We define the matching number $\nu(X,\tau)$ to be the smallest cardinal $\kappa$ such that for every matching ${\cal M}\subseteq \tau$ we have $|{\cal M}|\leq \kappa$.
Recall that $D\subseteq S$ is dense if $D$ intersects every non-empty open set. In the language of hypergraphs, dense sets are vertex covers. So we define the vertex covering number $t(X,\tau)$ to be the smallest cardinality that a dense set can have.
König's theorem that for all finite bipartite graphs $B$ we have $\nu(B)=\tau(B)$, when using the equivalents of the definitions above for graphs.
Let $(X,\tau)$ be a non-empty Hausdorff space.
Questions.
1) Do we always have $\nu(X,\tau)\leq t(X,\tau)$?
2) Do we always have $\nu(X,\tau)\geq t(X,\tau)$?
 A: What you are calling the "matching number" of $X$ is usually called its Souslin number -- the smallest cardinal bounding the size of any collection of pairwise disjoint open subsets of $X$.
What you are calling the "vertex covering number" of $X$ is usually called its density.
Thus your question is how the Souslin number and the density of a Hausdorff space are related. The answer is that the density is always $\geq$ the Souslin number, but the inequality can be strict. Thus the answer to your first question is yes, and the answer to the second question is no.
(Denote by $S(X)$ and $d(X)$ the Souslin number and density of $X$, respectively. To see that $S(X) \leq d(X)$, just observe that any dense subset $D$ of $X$ is at least as large as a collection $\mathcal U$ of pairwise disjoint open subsets of $X$, because $D$ must contain a point from each member of $\mathcal U$. To see that this inequality can be strict, try to show that the space $[0,1]^\kappa$ has countable Souslin number, but fails to be separable for any $\kappa > \mathfrak{c}$.)
