The second last implication $\{X_{n}\leq a\}\in \sigma(\mathscr{C})\Rightarrow \{X_{n}\leq a\}\subset \sigma(\tilde{X})$ is false, and I don't see how it is relevant here. One reason why this makes no sense is that $\{X_{n}\leq a\}$ is a subset of $\Omega$ and $\sigma(\tilde{X})$ is a subset of the power-set of $\Omega$. This also means that in start you should correct the definition of $\sigma(\tilde{X})$, the right way to write it would be $\sigma(\tilde{X})=\sigma(\{\{X_{n}\leq a\}:a\in\mathbb{R},\,\,n\in\mathbb{N}\})$. The last implications should instead be just:
\begin{align*}
&\{X_{n}\leq a\}\in\sigma(\mathscr{C}) \\
&\Rightarrow \sigma(\tilde{X})\subset \sigma(\mathscr{C}),
\end{align*}
otherwise it's correct. And this is the case because you just showed that $\{X_{n}\leq a\}\in \sigma(\mathscr{C})$ for all $a\in\mathbb{R}$ and $n\in\mathbb{N}$. Since $\sigma(\tilde{X})$ is the smallest $\sigma$-algebra that contains such sets and $\sigma(\mathscr{C})$ is some $\sigma$-algebra containing them, then $\sigma(\tilde{X})\subset \sigma(\mathscr{C})$.
Well, you "just" find another $\pi$-system. Note that $\sigma(X_1,...,X_n)$ is generated by sets of the form $(X_{j_1}\in B_1,...x_{j_k}\in B_k)$ for $(B_m)_{1\leq m\leq k}$ Borel sets and $1\leq j_1<...<j_k\leq n$, and the system of such sets is a $\pi$ system. Similarly, $\sigma(X_j)_{j>n}$ is generated by sets of the form $(X_{j_1}\in B_1,..., X_{j_k}\in B_k)$ again for a finite sequence of $(B_m)_{1\leq m\leq k}$ of Borel sets and $n<j_1<...<j_k$. Again this is a $\pi$ system.
Hence, fix Borel sets $(B_m)_{1\leq m\leq k}$ and $n<j_1<...j_k$ and let $(C_r)_{1\leq r\leq K}$ be any other finite sequence of Borel sets and $1\leq l_1<...<l_K\leq n$ be a finite increasing set of indices. Then, we see, by the independence assumption, that
$$
\mathbb{P}((\cap_{m=1}^k (X_{j_m}\in B_m))\cap (\cap_{r=1}^K X_{l_r}\in C_r))=\mathbb{P}(\cap_{m=1}^k (X_{j_m}\in B_m))\mathbb{P}(\cap_{r=1} (X_{l_r}\in C_r)),
$$
implying by Dynkin's lemma that the two measures on $\sigma((X_j)_{1\leq j\leq n})$ given by
$$
\mathbb{P}_1(A)=\mathbb{P}((\cap_{m=1}^k (X_{j_m}\in B_m))\cap A)
$$
and
$$
\mathbb{P}_2(A)=\mathbb{P}(\cap_{m=1}^k (X_{j_m}\in B_m))\mathbb{P}(A)
$$
are equal.
Since our choice of Borel sets $B_m$ and indices $j_m$ was arbitrary, this now yields, by another application of Dynkin's Lemma, that for any $A\in \sigma(X_j)_{1\leq j\leq n},$ the two measures on $\sigma(X_j)_{j>n}$ given by
$$
\mathbb{P}_3(D)=\mathbb{P}(D\cap A)
$$
and
$$
\mathbb{P}_4(D)=\mathbb{P}(D)\mathbb{P}(A)
$$
are equal.
However, since $A$ was arbitrary, this implies that the two $\sigma$-algebras are independent.
Best Answer
A very useful (if not the most useful) characterization of $\sigma$-algebra generated by random variables is the following theorem, paraphrased from Theorem 20.1 in Billingsley's Probability and Measure:
So, to answer your question, $\sigma\left(\sum_{i=1}^nX_i\right)\subseteq \sigma(X_1,\dots,X_n)$ since the summation is certainly a measurable function. And in fact, the above characterization allows you to quickly determine whether a random variable is in a generated $\sigma$-algebra in many cases.
There are too many different things you can say about $\sigma$-algebra generated by independent random variables. You may take a look at $\S$4, $\S14$, $\S$20 in Billingsley, or you can rephrase your question to be more specific.