Showing that $\|(x,y)\|_0=\sqrt{\|x\|^2+\|y\|^2}$ is norm if $\|\cdot\|$ is a norm. The 3 properties are really easy to show but I cannot show that $\|(x,y)\|_0=\sqrt{\|x\|^2+\|y\|^2}$ satisfies the triangle inequality if $\|\cdot\|$ satisfies it. I tried to use Cauchy,S. inequality and etc.
My work:
Let $(x,y),(x',y')\in X\times Y$, $X,Y$ are both normed spaces.
$$\|(x,y)+(x',y')\|_0=\|(x+x',y+y')\|_0=\sqrt{
\|x+x'\|^2+\|y+y'\|^2}\\\le\sqrt{(\|x\|+\|x'\|)^2+(\|y\|+\|y'\|)^2}$$
And I have shown that $\max\{\|x\|+\|x'\|,\|y\|+\|y'\|\}\le \max\{\|x\|,\|y\|\}+\max\{\|x'\|,\|y'\|\}$
But here I am stuck.
 A: Let $f: \mathbb{R}^n \to \mathbb{R}$ be a norm having the property that for any $v, w \in \mathbb{R}^n$ where the absolute value ($l_1$ norm) of any element of $v$ is at least the absolute value of the respective element of $w$, then $f(v) \geq f(w)$. Let us refer to this property as monotonicity. Let us also define $g: V = U_1 \times U_2 \times \cdots \times U_n \to \mathbb{R}^n$ is a function that returns a vector of $n$ elements where the $i$th element of the output is the norm of the $i$th input. Here all the $U_i$ are just arbitrary normed spaces. Then $f \circ g$ is a norm on $V$, as it satisfies

*

*Non-Negativity: The output of $f$ is always non-negative, so the output of $f \circ g$ is also non-negative.


*Positivity: Note that $f(u) = 0 \iff u = 0 \in \mathbb{R}^n$ and furthermore, $g(v) = 0$ is the case if and only if the $i$th element of $v$ was the zero element of $U_i$, for all $1 \leq i \leq n$. Hence, $$f(g(v)) = 0 \iff g(v) = 0 \iff v = 0 \in V$$


*Scalability: Note that $f(g(av)) = f(|a| \cdot g(v))$, since $g(v)$ is individually applied to each element of its input. We can then take $|a|$ out of $f$ because $f$ too is a norm.


*Triangle Inequality: Note that $g(u + v) \leq g(u) + g(v)$. But note that the elements of $g(u + v)$ and $g(u) + g(v)$ are all non-negative real numbers, and hence the absolute value of every member of $g(u + v)$ is at most the the absolute value of the respective member in $g(u) + g(v)$. Since $f$ is a monotonic norm by assumption, it follows that
$$f(g(u + v)) \leq f(g(u) + g(v)) \leq f(g(u)) + f(g(v))$$
as $f$ itself satisfies the triangle inequality since it is a norm.
So for your question, we can take $V = X \times Y$, take $f$ to be a function from $\mathbb{R}^2 \to \mathbb{R}$ with
$$f(x, y) = \sqrt{x^2 + y^2}$$
and $g$ to be the function from $V \to \mathbb{R}^2$ defined by
$$g(x, y) = (\|x\|, \|y\|)$$
and we will have our result.
Remark: As Daniel Fischer pointed out, not all norms on $\mathbb{R}^n$ are monotonic! See the comments for an example. However, note it is indeed the case that $f$ in your example is monotonic (it is just the $l_2$ norm).
A: As you have shown, $\|(x,y)+(x',y')\|_0^2=\|x+x'\|^2+\|y+y'\|^2$. Since $\|\cdot\|$ is a norm, we have 
$$\|(x,y)+(x',y')\|_0^2
\leq (\|x\|+\|x'\|)^2+(\|y\|+\|y'\|)^2\\
=\|x\|^2+\|x'\|^2+2\|x\| \|x'\|+\|y\|^2+\|y'\|^2+2\|y\| \|y'\|.$$
Since
$$\big(\|(x,y)\|+\|(x',y')\|\big)^2
=(\sqrt{\|x\|^2+\|y\|^2}+\sqrt{\|x'\|^2+\|y'\|^2})^2\\
=\|x\|^2+\|y\|^2+\|x'\|^2+\|y'\|^2
+2\sqrt{(\|x\|^2+\|y\|^2)(\|x'\|^2+\|y'\|^2)},$$
it suffices to prove
$$\|x\| \|x'\|+\|y\| \|y'\|\leq\sqrt{(\|x\|^2+\|y\|^2)(\|x'\|^2+\|y'\|^2)}.$$
Squaring it and doing some cancellation, we can see that it is equivalent to 
$$2\|x\| \|x'\|\|y\| \|y'\|\leq \|x\|^2\|y'\|^2+\|x'\|^2\|y\|^2,$$
which follows from the Cauchy-Schwarz inequality. 
