The answer is yes.
Given a cardinal $\kappa$, let $\lambda = 2^\kappa$ (this denotes cardinal exponentiation) and let $X_0 = 2^\lambda$ (this denotes a Tychonoff product). Clearly $|X_0| > \kappa$. $X$ has a dense subset of size $\kappa$, and it has the ccc, which means that every "matching" in $X_0$ is countable. (By the way, what you're calling a *matching* is usually called a *cellular family*.) If we take $X$ to be $\kappa$ disjoint copies of $X_0$ and set $\alpha = \kappa$, then all your conditions are met.

By the way, an interesting variant of this question is obtained by asking for $\alpha < \kappa$ instead of $\alpha < |X|$. In this version, the answer is yes for some $\kappa$ and no for others. (By the argument above, we get a yes answer if there is some $\alpha$ with $\alpha < \kappa \leq 2^\alpha$. But the answer is no if $\kappa$ is a strong limit cardinal, because if a Hausdorff space has a dense subset of size $\alpha$, then its cardinality is at most $2^{2^\alpha}$.)