Prove that $X$ is Hausdorff, if and only if for each pair $x, y \in X$, where $x \ne y$, there exists $j \in J$, such that $f_j (x) \ne f_j(y)$. 
Suppose that in the set $X$ we have the topology induced by the family $(f_j)_{j \in J}$, where $f_j : X \to Y_j$ and the spaces $Y_j$ are Hausdorff spaces. Prove that $X$ is Hausdorff, if and only if for each pair $x, y \in X$, where $x \ne y$, there exists $j \in J$, such that $f_j (x) \ne f_j(y)$.

Is it ok to approach the problem with direct proof?
Suppose that $X$ is Hausdorff and let distinct $x,y \in X$. Then by the property of Hausdorff spaces we have neighborhoods $O_x$ and $O_y$ for which $O_x \cap O_y = \emptyset$. Now since $X$ has the induced topology we have that $$x \in \bigcap_{j \in F} f_j^{-1}(U_j) \subset O_x \\ y \in \bigcap_{j \in F} f_j^{-1}(V_j) \subset O_y$$ for finite $F$ and $U_j, V_j$ open in $Y_j$. Thus for all $j \in F$ we have that $x \in f_j^{-1}(U_j) \implies f_j(x) \in U_j$ and $y \in f_j^{-1}(V_j) \implies f_j(y) \in V_j$. Because each $Y_j$ is Hausdorff we have that $U_j \cap V_j = \emptyset$ for all $j$ so it must be that $f_j(x)\ne f_j(y)$ since they're contained in distinct neighborhoods?
 A: Your direct proof doesn't quite work for the forward direction you sketch. If $X$ is Hausdorff and $x \neq y$, we do have basic open $O_x$ and $O_y$ that are disjoint neighbourhoods of $x$, and $y$ resp.
For $O_x$ we can write $O_x= \bigcap_{j \in F_1} f_j^{-1}[U_j]$ for some finite subset $F_1 \subset J$ and all $U_j$ open in the corresponding space $Y_j$. Similarly,  $O_x= \bigcap_{j \in F_2} f_j^{-1}[V_j]$ for some finite subset $F_2 \subset J$ and all $V_j$ open in the corresponding space $Y_j$.
But $F_1 \cap F_2 = \emptyset$ is possible. You suggest we have then very same $F$ for both. We could sort of fix that in the disjoint case by setting $V_j = Y_j$ for $j \in F_1$ and also $U_j=Y_j$ for $j \in F_2$ and add extra $X$'s in the intersection (which have no effect on the intersection) and then use $F = F_1 \cup F_2$). But then the rest of the supposed proof doesn't work: whence the claim that always $U_j\cap V_j = \emptyset$ by Hausdorffness? It's nonsense, the $f_j$ need not be injective at all.
The indirect proof does work: suppose the property fails then there are $x \neq y$ in $X$ so that $f_j(x)=f_j(y)$ for all $j$. Then for any subbasic set $B=f_j^{-1}[O]$ we always have "$x \in B$ iff $y \in B$" and this equivalence extends trivially to the generated base as well, and hence $X$ is not even $T_0$, let alone Hausdorff. Far easier and less notation too.
