$X_1,\dots,X_n$ being independent is, by definition, equivalent to $\sigma(X_1),\dots,\sigma(X_n)$ being independent which is further defined to mean that
$$
P(A_1\cap\cdots\cap A_n)=P(A_1)\cdots P(A_n) \quad\text{ for all }A_i\in \sigma(X_i)
$$
Now, since $\sigma(X_i)$ consists precisely of sets of the form $(X_i\in B_i)$ where $B_i\in{\cal B}(\mathbb{R})$, the definition of independence can be expressed equivalently as
$$
(1) \qquad P(X_1\in B_1,\dots ,X_n\in B_n)=P(X_1\in B_1)\dots P(X_n\in B_n) \quad\text{ for all }B_i\in {\cal B}(\mathbb{R})
$$
Next, in (1), take $B_1=\cdots=B_k=\Omega$. Then (1) becomes
$$
P(X_1\in \Omega,\dots,X_k\in\Omega,X_{k+1}\in B_{k+1},\dots,X_n\in B_n)
\\
=P(X_1\in \Omega)\cdots P(X_k\in\Omega)P(X_{k+1}\in B_{k+1})\cdots P(X_n\in B_n)
$$
which simplifies to
$$
(2)\qquad P(X_{k+1}\in B_{k+1},\dots,X_n\in B_n)
=P(X_{k+1}\in B_{k+1})\cdots P(X_n\in B_n)
$$
A similar argument gives
$$
(3)\qquad P(X_1\in B_1,\dots,X_k\in B_k)
=P(X_1\in B_1)\cdots P(X_k\in B_k)
$$
Hence, substituting (2) and (3) into (1) gives
\begin{eqnarray*}
&&P((X_1,\dots,X_k,X_{k+1},\dots,X_n)\in B_1\times\cdots\times B_k\times B_{k+1}\times\cdots\times B_n)\\
&=&P(X_1\in B_1,\dots,X_k\in B_k,X_{k+1}\in B_{k+1},\dots,X_n\in B_n)\\
&=&
P(X_1\in B_1)\cdots P(X_k\in B_k)P(X_{k+1}\in B_{k+1})\cdots P(X_n\in B_n)\\
&=&
\Big[P(X_1\in B_1)\cdots P(X_k\in B_k)\Big]\Big[P(X_{k+1}\in B_{k+1})\cdots P(X_n\in B_n)\Big]\\
&=&
\Big[P(X_1\in B_1,\dots,X_k\in B_k)\Big]\Big[P(X_{k+1}\in B_{k+1},\dots,X_n\in B_n)\Big]\\
&=&
\Big[P((X_1,\dots,X_k)\in B_1\times\cdots\times B_k)\Big]\Big[P((X_{k+1},\dots,X_n)\in B_{k+1}\times\cdots\times B_n)\Big]\\
\end{eqnarray*}
which holds for all $B_i\in {\cal B}(\mathbb{R})$.
Now, sets of the form $B_1\times\cdots\times B_k$, where the $B's\in{\cal B}(\mathbb{R})$, generate ${\cal B}(\mathbb{R}^k)$. Hence, sets of the form $((X_1,\dots,X_k)\in B_1\times\cdots\times B_k)$ generate $\sigma(X_1,\dots,X_k)$. Similarly, sets of the form $B_{k+1}\times\cdots\times B_n$, where the $B's\in{\cal B}(\mathbb{R})$, generate ${\cal B}(\mathbb{R}^{n-k})$. And sets of the form $((X_{k+1},\dots,X_n)\in B_{k+1}\times\cdots\times B_n)$ generate $\sigma(X_{k+1},\dots,X_n)$. Hence, for any $M\in{\cal B}(\mathbb{R}^k)$ and $N\in{\cal B}(\mathbb{R}^{n-k})$ we have
\begin{eqnarray*}
&&P((X_1,\dots,X_k)\in M, (X_{k+1},\dots,X_n)\in N )\\
&=&P((X_1,\dots,X_k,X_{k+1},\dots,X_n)\in M\times N )\\
&=&\Big[P((X_1,\dots,X_k)\in M)\Big]\Big[P((X_{k+1},\dots,X_n)\in N)\Big]\\
\end{eqnarray*}
which shows that $(X_1,\dots,X_k)$ and $(X_{k+1},\dots,X_n)$ are independent.
Best Answer
Let $A_{12}$ and $A_1$ be any two Borel measurable sets, let also $\mu_1$, $\mu_2$ and $\mu_3$ be the respective probability measures of $X_1$, $X_2$ and $X_3$. Since the latter are independent, then any joint probability measure of those would be the product and so we can do the Lebesgue integration
\begin{align*} \mathbb{P}(X_1 + X_2 \in A_{12},X_3\in A_3) &= \int \int\int \mathbb{1}(x_1 + x_2 \in A_{12},x_3\in A_3)d\mu_1d\mu_2d\mu_3\\ &=\int \int\int \mathbb{1}(x_1 + x_2 \in A_{12})\mathbb{1}(x_3\in A_3)d\mu_1 d\mu_2 d\mu_3\\ &=\int\int\mathbb{1}(x_1 + x_2 \in A_{12}) d\mu_1 d\mu_2 \int \mathbb{1}(x_3\in A_3)d\mu_3\\ &=\mathbb{P}(X_1+X_2\in A_{12}) \mathbb{P}(X_3\in A_3) \end{align*}
So yes $X_1+X_2$ and $X_3$ are independent.
Let's try without the Lebesgue integration, we can show that $Y=(X_1, X_2)$ is independent of $X_3$ since for any Borel mesurable sets $A_1$, $A_2$, $A_3$ \begin{align*} \mathbb{P}(Y\in A_1\times A_2,X_3\in A_3)&=\mathbb{P}(X_1\in A_1, X_2\in A_2, X_3 \in A_3)\\ &=\mathbb{P}(X_1\in A_1, X_2\in A_2) \mathbb{P}(X_3\in A_3)\\ &=\mathbb{P}(Y\in A_1\times A_2) \mathbb{P}(X_3\in A_3) \end{align*}
Now let $f$ be any deterministic function over the domain of $Y$ and $f^{-1}(Z)=\lbrace y | f(y)\in Z\rbrace$, then for Borel measurable sets $A_0$, $A_3$ \begin{align*} \mathbb{P}(f(Y)\in A_0,X_3\in A_3)&=\mathbb{P}(Y\in f^{-1}(A_0),X_3\in A_3)\\ &=\mathbb{P}(Y\in f^{-1}(A_0))\mathbb{P}(X_3\in A_3)\\ &=\mathbb{P}(f(Y)\in A_0)\mathbb{P}(X_3\in A_3)\\ \end{align*}
So $f(Y)$ is independent of $X_3$.