The answer is no. E.g., let $x_{2n}=x:=(\frac12,\frac12,\frac13,\frac14,\ldots)$ and $x_{2n+1}=y:=(0,\frac12,\frac13,\frac14,\ldots)$ (for all $n$). Then (i) $x_{2n}=x\to x$ and $x_{2n+1}=y\to y$ in any sense and (ii) $\|x_{2n}+x_{2n+1}\|=2\|x_{2n}\|=2\|x_{2n+1}\|$ and hence $I_n=0$ for all. However, $x\ne y$. $\quad\Box$

The answer is no. E.g., let $x_{2n}=x:=(\frac12,\frac12,\frac13,\frac14,\ldots)$ and $x_{2n+1}=y:=(0,\frac12,\frac13,\frac14,\ldots)$ (for all $n$). Then (i) $x_{2n}=x\to x$ and $x_{2n+1}=y\to y$ in any sense and (ii) $\|x_{2n}+x_{2n+1}\|=2\|x_{2n}\|=2\|x_{2n+1}\|$ and hence $I_n=0$ for all $n$. However, $x\ne y$. $\quad\Box$

The answer is no. E.g., let $x:=(\frac12,\frac12,\frac13,\frac14,\ldots)$ and $y:=(0,\frac12,\frac13,\frac14,\ldots)$. Then $\|x+y\|=\|x\|+\|y\|$, but $x$ and $y$ are linearly independent.

  However, it is rather easy to see that the answer will become yes if the summation $\sum_{i=2}^{\infty}$ in the definition of the norm is replaced by $\sum_{i=1}^{\infty}$.

The answer is no. E.g., let $x_{2n}=x:=(\frac12,\frac12,\frac13,\frac14,\ldots)$ and $x_{2n+1}=y:=(0,\frac12,\frac13,\frac14,\ldots)$ (for all $n$). Then (i) $x_{2n}=x\to x$ and $x_{2n+1}=y\to y$ in any sense and (ii) $\|x_{2n}+x_{2n+1}\|=2\|x_{2n}\|=2\|x_{2n+1}\|$ and hence $I_n=0$ for all. However, $x\ne y$. $\quad\Box$