Does this prove continuity? Suppose I have a function $f\colon X\times Y\to Z$ and suppose on the product space $X\times Y$ I have a metric $d$ which is a sum of two metrics, say
$$
d((x,y),(x',y'))=d_1(x,x')+d_2(y,y').
$$
On $Z$, I have metric $d_Z$.
Now, if I want to show that $f$ is continuous, I have to show that for each $\varepsilon >0$ there exists some $\delta>0$ such that
$$
d((x,x'),y,y'))\leq\delta\implies d_Z(f(x,y),f(x',y'))\leq\varepsilon.
$$
Now, my question is if the following is appropriate or nonsens:
Does the following show continuity of $f$?
For each $\varepsilon >0$, if I set $y=y'$, i.e. $d_2(y,y')=0$, I can find a $\delta>0$ such that the implication above holds. 
For me this is strange since I always have to have $d_2(y,y')=0$. I do not know if this really proves continuity then.
 A: *

*In order to be able to say that $f$ is continuous at $(x_0,y_0)$ you should show that for any $(x,y)$ satisfying $d((x_0,y_0),(x,y)) \leq \delta $ the values of $f$ should be close i.e. $d_z(f(x,y), f(x_0,y_0)) \leq \epsilon $. In that sense, if you pick $y=y_0$ then you are not working with arbitrary points anymore. 

*Consider the following: $X,Y= \mathbb{R}$ with the standard metric. Then $Z= \mathbb{R}^2$ with taxicab metric. Now consider $f(x,y) = 1 $ if $y \in \mathbb{Q}$ and $f(x,y)= 0$ otherwise. In this case, $f$ is not continuous at any point but continuous on horizontal lines, i.e. when $y=y_0$ fixed.

A: No, what you did does NOT prove continuity, so you're right in feeling doubtful about this. The reason for why it's wrong can be clearly seen if you set up the actual $\varepsilon$-$\delta$ definition of continuity:

$f:A\to B$ is continuous at a point $a_0$ if for any $\varepsilon>0$ there exists $\delta>0$ such that for any $a\in A$ we have: $d(a,a_0)<\delta \implies d(f(a),f(a_0))<\varepsilon$.

Applied to the situation in your question, this becomes:

$f:X\times Y\to Z$ is continuous at a point $(x,y)$ if for any $\varepsilon>0$ there exists $\delta>0$ such that for any $(x',y')\in X\times Y$ we have: $d((x,y),(x',y'))<\delta \implies d_Z(f(x,y),f(x',y'))<\varepsilon$.

In this definition, finding an appropriate $\delta$ comes before considering points $(x',y')$, and moreover the condition must hold for all such points — meaning you can't select it in some special convenient way.
Your mistake was that you chose some special points $(x',y')$ (which you can't do) and you chose them before defining $\delta$ (which you can't do either, as $\delta$ must be defined first). What you did would effectively demonstrate continuity of $f$ as a function of a single variable $x$ only while holding $y$ constant. But that's not the same as proving the (joint) continuity of $f$.
And one more comment. The definition that you stated in your post is in fact the definition of a stronger property of uniform continuity:

$f:X\times Y\to Z$ is uniformly continuous if for any $\varepsilon>0$ there exists $\delta>0$ such that for any $(x,y),(x',y')\in X\times Y$ we have: $d((x,y),(x',y'))<\delta \implies d_Z(f(x,y),f(x',y'))<\varepsilon$.

