Show that $\phi_1:X\times Y \to X$ given by $\phi_1(x,y):=x$ is continuous on metric spaces 
Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. Then show that the function $\phi_1:X\times Y \to X$ given by
$$\phi_1(x,y)=x$$
is $(d,d_x)$-continuous. With the fact that $d((x_1, y_1), (x_2, y_2))=d_X(x_1,x_2)+d_Y(y_1,y_2)$.

So far, I have got the using the definition of continuous functions in metric spaces that there must exist an $\epsilon>0$, then there exists a $\delta>0$ such that $$d(x,y) \lt \delta \implies d_X(\phi(x), \phi(y)) \lt \epsilon.$$ But I don't understand how to use the fact given, and how do you know what $d_X$ is?
Would this be correct:
$$d((x_1,y_1),(x_2,y_2)) \lt \delta \implies d_X(x_1, x_2) \lt \epsilon$$
then taking $\delta= \epsilon$, then
$$d_X(x_1,x_2) \leq d((x_1,y_1),(x_2,y_2)) =d_X(x_1,x_2)+d_Y(y_1,y_2) \lt \delta = \epsilon.$$
Does this satisfy that it is continuous over the metric space?
 A: Your use of the definition of continuity in metric spaces is a bit off. There is no 'there exists $\epsilon>0$', it is '... given $\epsilon>0$, there exists $\delta>0$ such that ...' In full:

If $(X,d_X)$ and $(Y,d_Y)$ are metric spaces, then we say that $f:X\to Y$ is $(d_X,d_Y)$-continuous at $x_0\in X$ if for each given $\epsilon>0$, there exists $\delta>0$ such that for all $x\in X$ with $d_X(x,x_0)<\delta$, we have that $d_Y(f(x),f(x_0))<\epsilon$. We say that $f$ is $(d_X,d_Y)$-continuous (on the whole of $X$) if it is $(d_X,d_Y)$-continuous at each point $x_0\in X$.

Usually we drop the '$(d_X,d_Y)$' when the metrics involved are clear from the context.
Here is how I would prove it:

Proposition: Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and define the metric $$d((x_1,y_1),(x_2,y_2)):=d_X(x_1,x_2)+d_Y(y_1,y_2)$$ on $X\times Y$. Then $\phi_1:X\times Y\to X$ given by $\phi_1(x,y):=x$ is $(d,d_X)$-continuous.

Proof: Let $(x_0,y_0)\in X\times Y$ and $\epsilon>0$. Then for each $(x,y)\in X\times Y$ with $$d((x,y),(x_0,y_0))<\epsilon,$$ we have that $$d_X(\phi_1(x,y),\phi_1(x_0,y_0))=d_X(x,x_0)\leq d_X(x,x_0)+d_Y(y,y_0)=d((x,y),(x_0,y_0))<\epsilon.$$ This shows that taking $\delta=\epsilon$ works. Since $\epsilon>0$ was chosen arbitrarily, we have shown directly from the definition of continuity above that $\phi_1$ is continuous at $(x_0,y_0)$ (for each $(x_0,y_0)\in X\times Y$), and we are done.
A: see that
$$d_X(\phi_1(x,y), \phi_1(z,w)) = d_X(x,z)\leq d_X(x,z)+d_Y(y,w) = d((x,y),(z,w))$$
So $\phi$ is Lipschitz with respect to the given metrics, so of course it's continuous. in this case, just actually take $\epsilon=\delta$ .
