Any collection of neighborhoods of a dense set cover the space?

This is false even in the real line. Let $\{r_1,r_2,...\}$ be an ennumeration of the rationals and consider the neighborhoods $(r_n-\frac 1 {2^{n}},r_n+\frac 1 {2^{n}})$. Since the total length of these intervals is $\sum_{n=1}^{\infty} \frac 1 {2^{n-1}}=2$, they cannot cover the entire line. [ If you know measure theory you can see the reason immediately. Otherwise you can see that the intervals don't even cover $[0,3]$ using a compactness argument.]

If $y\notin D$ then you can choose for $U_x=X-\{y\}$ for every $x\in D$ - which is evidently an open neighborhood of $x$ in Hausdorff space $X$.

Then the union of these sets is again $X-\{y\}\neq X$.

Apparantly that statement is false already if $X$ is a $T_1$-space.