Recall that a topological space $(X,\tau)$, where $\tau$ denotes the collection of open subsets of $X$, is Hausdorff if for any two distinct points $x,y\in X$, there exist open sets $U_x,U_y\in\tau$ such that $U_x\cap U_y=\varnothing$, $x\in U_x$, and $y\in U_y$.
Now as for the particular example: Pick any two points $x,y\in\mathbb R$. The ordinary Euclidean topology on $\mathbb R$ is Hausdorff. Therefore, there exist disjoint open sets $U_x$ and $U_y$ such that $x\in U_x$ and $y\in U_y$. Now note that $U_x=U_x\cup(\varnothing\cap\mathbb Q)\in T$, because the empty set is open in the usual topology. Similarly, $U_y\in T$. Therefore, $U_x$ and $U_y$ are open in $T$ as well, which implies that $T$, too, is a Hausdorff topology.
You should also check that the union of arbitrarily many members of $T$ is also in $T$. To this end, suppose that $\{W_{\alpha}\}_{\alpha\in A}$ is a collection of $T$-open sets, where $A$ is a non-empty index set (may be finite, countably infinite, or uncountably infinite). Since $W_{\alpha}\in T$ for each $\alpha\in A$, $W_{\alpha}=U_{\alpha}\cup(V_{\alpha}\cap\mathbb Q)$ for some $U_{\alpha},V_{\alpha}$ open in $\mathbb R$. Then, it is not difficult to check that
$$\bigcup_{\alpha\in A} W_{\alpha}=\left(\bigcup_{\alpha\in A} U_{\alpha}\right)\cup\left[\left(\bigcup_{\alpha\in A} V_{\alpha}\right)\cap\mathbb Q\right].$$
Now, $\bigcup_{\alpha\in A} U_{\alpha}$ and $\bigcup_{\alpha\in A} V_{\alpha}$ are open in $\mathbb R$, which implies that $\bigcup_{\alpha\in A} W_{\alpha}\in T$.
[Warning: In general, it is not sufficient to check this property for the union of two sets. It is logically invalid to extend the case of the union of two sets to the union of to infinitely many sets; you should be working with infinitely many sets from the outset. It is valid, however, to extend the two-set result to the case of the union of finitely many sets: just use induction to prove this.]
As for finite intersections, it is sufficient to check the case of the intersection of only two sets, and the general finite case follows easily by induction (cf. the parenthesized comment above). So let $W_1,W_2\in T$. Then
\begin{align*}
W_1=&\,U_1\cup(V_1\cap\mathbb Q),\\
W_2=&\,U_2\cup(V_2\cap\mathbb Q),
\end{align*}
where $U_1,U_2,V_1,V_2$ are open in $\mathbb R$. Then,
\begin{align*}
W_1\cap W_2=&\,(U_1\cap U_2)\cup(U_1\cap V_2\cap \mathbb Q)\cup(U_2\cap V_1\cap \mathbb Q)\cup(V_1\cap V_2\cap\mathbb Q)\\
=&\,\underbrace{(U_1\cap U_2)}_{\text{open in $\mathbb R$}}\cup\{\underbrace{\left[(U_1\cap V_2)\cup(U_2\cap V_1)\cup(V_1\cap V_2)\right]}_{\text{open in $\mathbb R$}}\cap\mathbb Q\}.
\end{align*}
(I leave it to you to check whether the set algebra is correct.) Hence, $W_1\cap W_2\in T$ because it can be represented in the desired form.
Finally, as per the definition of a topological space, you should also check that $\varnothing\in T$ and $\mathbb R\in T$. This is an easy exercise.
