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?
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
$$
X^{-1}(\mathcal{B})\subseteq\sigma(X^{-1}(\mathcal{U})).
$$
Best Answer
By definition, $\sigma(Z)$ denotes the smallest $\sigma$-algebra $\Sigma$ on $\Omega$ such that $Z: (\Omega, \Sigma) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ is measurable.
This means in particular that $X:(\Omega,\sigma(X)) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ is measurable. Since $\sigma(Y) = \sigma(X)$, we also have that
$$Y: (\Omega,\sigma(X)) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$$
is measurable. Consequently, the sum
$$X+Y: (\Omega,\sigma(X)) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$$
is measurable as a sum of two measurable random variables. Since $\sigma(X+Y)$ is the smallest $\sigma$-algebra $\Sigma$ such that
$$X+Y: (\Omega,\Sigma) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$$
is measurable, this shows $\sigma(X+Y) \subseteq \sigma(X)$.