The sigma-algebra generated by $1_{[0,1/2]}$ is simply
$$
\bigl\{\emptyset,[0,1],[0,1/2],(1/2,1]\bigr\}.
$$
It consists of the preimages under the function $1_{[0,1/2]}$ of all Borel sets in the codomain of the function $1_{[0,1/2]}$, namely, $(\mathbb R,B(\mathbb R))$. (Notice that the preimage $1_{[0,1/2]}^{-1}(M)$ is completely determined by the information of whether 0 and 1 do or do not belong to $M$ respectively.)
The situation for $1_{[1/4,3/4]}$ is similar.
The random variables $1_{[0,1/2]}$ and $1_{[1/4,3/4]}$ on $([0,1],B[0,1],L)$ are indeed independent: For this you have to check that $L(A\cap B)=L(A)\cdot L(B)$ for all $A\in 1_{[0,1/2]}^{-1}(B(\mathbb R))$ and $B\in 1_{[1/4,3/4]}^{-1}(B(\mathbb R))$.
The most interesting case is $L([0,1/2]\cap [1/4,3/4])=L([0,1/2])\cdot L([1/4,3/4])$.
Check that both sides are equal!
Also think about the following question: Are the random variables $1_{[0,1/2]}$ and $1_{[1/4,1]}$ on $([0,1],B[0,1],L)$ also independent?
The definition says that $X$ is $F$-measureable, which (as Alex pointed out) amounts to requiring the sublevel sets $\{\omega:X(\omega)\le \alpha\}$ to be in $F$ for every real number $\alpha$. (Equivalently, one can ask for this property with $<\alpha$, $\ge\alpha$, or $>\alpha$).
To see why we might care about such a thing, consider the special case of $X$ taking values $0$ and $1$ only. Then the expected value of $X$ is the measure of the set $\{\omega:X(\omega)=1\}$ which is not defined unless this set is in $F$. In general, measurability is needed to compute the expected value, variance, and other quantities describing $X$.
Studying the subject further, one finds that measurability can also express the idea of a random variable being independent of something.
Best Answer
Let $\mathcal{U}=\{U\subseteq\mathbb{R}\mid U\text{ is open}\}$ denote the open subsets of $\mathbb{R}$ and $\mathcal{B}$ denote the Borel sets of $\mathbb{R}$. Then we want to show that $X^{-1}(\mathcal{B})\subseteq \sigma(X^{-1}(\mathcal{U}))$.
To that end show that $$\mathcal{A}:=\{B\subseteq\mathbb{R}\mid X^{-1}(B)\in \sigma(X^{-1}(\mathcal{U}))\}$$ is a sigma-algebra. Then use the fact that $\mathcal{U}\subseteq\mathcal{A}$ to conlude that $\sigma(\mathcal{U})\subseteq\mathcal{A}$. Now this means that $$ X^{-1}(B)\in\sigma(X^{-1}(\mathcal{U})),\quad \text{for all}\,\, B\in\sigma(\mathcal{U}), $$ or in other words $$ X^{-1}(\mathcal{B})\subseteq\sigma(X^{-1}(\mathcal{U})). $$