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I'm pretty sure it's not closed under vector addition but verification would be appreciated.

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If by $x1$ you mean the first coordinate of $x$, then the set consists actually of those vectors whose all other coordinates except possibly the first are zero. – J. J. Oct 18 '11 at 6:06
Look at the case $n=2$ for concreteness. Look at the vector $(a,b)$. To say its norm is $|a|$ is to say that $a^2+b^2=a^2$, so it's a fancy way of saying $b=0$. For sure it is a subspace! Argument for general $n$ is the same. – André Nicolas Oct 18 '11 at 6:07

Notice $$\sqrt{x_1^2+x_2^2+\cdots+x_n^2}=|x_1|\implies x_1^2+x_2^2+x_3^2+\cdots+x_n^2=|x_1|^2.$$ But $x_1^2=|x_1|^2$, so subtracting gives $x_2^2+x_3^2+\cdots+x_n^2=0$. This is only possible if all but the first component of the vector is zero (can you see why?). Also, $r \vec{e}_1\in S$ for any real $r\in\mathbb{R}$ and $\vec{e}_1$ the unit vector on the $x_1$-axis (check this!), so we may conclude $$S=\{(r,0,0,\dots,0)\in\mathbb{R}^n:r\in\mathbb{R}\}.$$ Can you tell whether or not this is a vector space in its own right?

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