$S:\mathbb{R}^2 \to \mathbb{R}$ , $S(x,y)=x+y$ is continuous. Prove: $S:\mathbb{R}^2 \to \mathbb{R}$ , $S(x,y)=x+y$ is continuous.
My attempt:
$d((x,y),(a,b)) = d(x,a) + d(y,b)$
Let $\epsilon >0$.
I need to prove:
$d(x+y,a+b) < \epsilon$. If the following inequality is valid:
$d(x+y,a+b) \leq d(x,a) + d(y,b)$, then I can choose $\delta=\epsilon$.
But I can't verify that inequality.
Can someone please show me how to verify or if it is false show a counter-example?
I'm starting to think that this proof isn't that simple, probably the inequality I used is false. 
Thanks.
EDIT:
Please verify if this is correct:
Let $\delta = \epsilon$
$d(x+y,a+b) = |x+y-(a+b)| = |x+y-a-b| \leq |x-a| + |y-b| < \delta = \epsilon$
 A: It almost seems like you have it backwards. You should be fixing $\epsilon$ first and then finding a $\delta$ which is sufficiently small. You have correctly come to the inequality:
$$d(S(x,y),S(a,b))=d(x+y,a+b) \leq |x-a| + |y-b|$$
What we want to ensure is that for a given $\epsilon$, there exists a sufficiently small $\delta$. In other words, for any chosen $\epsilon$, can we choose a $\delta$ such that 
$$d((x,y),(a,b)) < \delta \quad \Rightarrow \quad |x-a| + |y-b| < \epsilon \quad ?$$
For a start, let's evaluate $d((x,y),(a,b))$:
$$d((x,y),(a,b)) = \sqrt{(x-a)^2 + (a-b)^2}$$
Note that this is greater than or equal to either $|x-a|$ or $|x-b|$. So
$$2 \cdot d((x,y),(a,b)) \geq |x-a| + |y-b|$$
$$2 \cdot d((x,y),(a,b)) \geq d(S(x,y),S(a,b))$$
Which means that for a given $\epsilon$, we can ensure that the right hand side is strictly less than it by setting
$$d((x,y),(a,b)) < \epsilon / 2$$
And so for a given $\epsilon$, the largest $\delta$ which will work in general is $\epsilon / 2$.
