Intuitively, the statement in the question body is true.
My definition of independence:
Let $X_1, \dots, X_k$ be random variables. Then they are called independent if for all Borel measurable sets $A_1, \dots, A_k$, we have
$$\mathbb{P}\{X_1 \in A_1, \dots ,X_k \in A_k\} = \prod_{i=1}^k \mathbb{P}\{X_i \in A_i\}$$
I tried to prove the result with this definition but I can't detach the two variables from the sum, which makes me unsure that the claim is even true.
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 $X_3$.
