Let me denote $x_{(1)}$ the largest element of $x$ in absolute value and $x_{(2)}$ the second one. Then you want to prove that $[(x+y)_{(1)}^2+(x+y)_{(2)}^2]^{1/2}\leq[x_{(1)}^2+x_{(2)}^2]^{1/2}+[y_{(1)}^2+y_{(2)}^2]^{1/2}$, in the square:
$(x+y)_{(1)}^2+(x+y)_{(2)}^2\leq x_{(1)}^2+x_{(2)}^2+y_{(1)}^2+y_{(2)}^2+2\sqrt{x_{(1)}^2y_{(1)}^2+x_{(1)}^2y_{(2)}^2+x{(2)}^2y_{(1)}^2+x_{(2)}^2y_{(2)}^2}$
There are two possibilities. First: $(x+y)_{(1)}=x_{(1)}+y_{(1)}$, then necesserily $(x+y)_{(2)}\leq x_{(2)}+y_{(2)}$. Second: $(x+y)_{(1)}, then necessarily (WLOG) $(x+y)_{(1)}\leq x_{(1)}+y_{(2)}$ and $(x+y)_{(2)}\leq x_{(2)}+y_{(1)}$ (it can happen that $x$ and $y$ are to be exchnged here).
Together: $(x+y)_{(1)}^2+(x+y)_{(2)}^2\leq x_{(1)}^2+x_{(2)}^2+y_{(1)}^2+y_{(2)}^2+2A+2B$, where $A+B=x_{(1)}y_{(1)}+x_{(2)}y_{(2)}$ or $A+B=x_{(1)}y_{(2)}+x_{(2)}y_{(1)}$. Both possibilities easily give $A+B\leq\sqrt{x_{(1)}^2y_{(1)}^2+x_{(1)}^2y_{(2)}^2+x_{(2)}^2y_{(1)}^2+x_{(2)}^2y_{(2)}^2}$.
Q.E.D.