Let $(A,d), (X,d')$ and $(Y,d'')$ be metric spaces and $f:A \to X$ and $g:A \to Y$. Also let $(X \times Y, e)$ be a metric space where $e(x,y)= \max\{d'(x,x'), d''(y,y')\}$ when $x=(x,y), y=(x',y').$ Show that the map $F:A \to X \times Y$, $F(a)=(f(a),g(a))$ is uniformly continuous iff $f,g$ are both uniformly continuous.
Here is my current attempts in both directions
$"\Rightarrow"$ Assume that $F$ is uniformly continuous then $d(x,y) < \delta \implies e(F(x),F(y)) = \max\{d'(f(x),f(y)), d''(g(x),g(y))\} < \varepsilon.$ The problem here is that this implies that either $d'(f(x),f(y)) < \varepsilon$ or $d''(g(x),g(y)) < \varepsilon$, but not both? The objective is to prove that both $f$ and $g$ are uniformly continous it seems that I can only get other one of these?
$"\Leftarrow"$ Assume $f,g$ are both uniformly continuous. Now $d(x,y) < \delta_f \implies d'(f(x),f(y)) < \varepsilon$ and $d(x,y) < \delta_g \implies d''(g(x),g(y)) < \varepsilon.$ Pick $\delta = \min\{\delta_f,\delta_g\}$. So now $d'(f(x),f(y)) < \varepsilon$ and $d''(g(x),g(y)) < \varepsilon$, when $d(x,y) < \delta$. From here $$e(F(x),F(y)) = \max\{d'(f(x),f(y)), d''(g(x),g(y))\}$$ and both these are less than $\delta$, but I'm not sure how to work with the $\max$ to show this?
Because for all elements $a,b,c$ in a linearly ordered set we have
$$\max(a,b) < c \iff a < c \land b < c$$
Left to right: $a \le \max(a,b) < c$ so $a < c$ and also $b \le \max(a,b) < c$ so $b < c$.
Right to left: If $a < c$ and $b < c$ then if $a \le b$ then $\max(a,b)=b < c$ and otherwise $b < a$ and then $\max(a,b)=a < c$ as well.