Show that for two subspaces $S,T$ of a vector space $V$, $S+T$ is direct iff $A,B$ are disjoint for all bases $A$ of $S$ and $B$ of $T$.
I am not so familiar with direct sums but all I know that $S+T$ is direct iff $S \cap T =${$0$}.Can this fact be used in this?
Yes, knowing that fact(definition) is enough, at least for this question.
(1) Assume that $S\cap T\neq\{0\}$. Take a basis $\gamma$ of $S\cap T$. We can extend it to a basis$^{[1]}$ $\alpha$ of $S$ and $\beta$ of $T$, resp. Then, $\alpha\cup\beta$ is a basis of $S+T^{[2]}$. (so we constructed $A,B$ : bases of $S,T$, resp. s.t. $A\cap B\neq\emptyset$)
(2) Assume that $S\cap T=\{0\}$. Assume that $A,B$ are bases of $S,T$, resp, with $v\in A\cap B$. Then, it means $v\in S\cap T$.
*[1] : you can show by an argument similar with showing existence of basis(using Zorn's lemma)
*[2] : It spans $S+T$ since $S+T=\{s+t|s\in S,t\in T\}$ and we can express such $s,t$ using $\alpha,\beta$. $\alpha\cup\beta$ is linearly independent for the following reason.
Assume we have $c_1\alpha_1+...+c_m\alpha_m+d_1\beta_1+...d_n\beta_n=0$, where $\alpha_1,...\alpha_m\in\alpha\setminus\beta$ and $\beta_1,...,\beta_n\in\beta$. $c_1\alpha_1+...c_m\alpha_m=-d_1\beta_1-...d_n\beta_n\in S\cap T$ so $c_1\alpha_1+...c_m\alpha_m$ is expressible using $\alpha\cap\beta$ and is uniquely expressible using $\alpha$, thus $c_1=...c_m=0$. Now remember that $\beta$ is a basis of $T$.