Uniformly equicontinuous

Let $X$ and $Y$ be metric space and $a \in X$. A family $A$ of functions from $X$ to $Y$ is said to be equicontinuous at $a$ if for any $\epsilon >0$ there exists a $\delta >0$ such that $d(x,a)< \delta$ implies $d(f(x), f(y) ) < \epsilon$ for all $f$ in $A$.

We say that $A$ is uniformly equicontinuous on $X$ if for any $\epsilon >0$ there exists $\delta >0$such that $d(x_1,x_2) < \delta$ implies $d(f(x_1), f(x_2) ) < \epsilon$ for all $f\in A$

Now let $X$ be acompact metric space. Let $F: X\times X \rightarrow Z$ be continuous. Let $f_y(x) = F(x,y)$. Show that $A = \{ f_y | y \in X \}$ is uniformly equicontinuous.

$X \times X$ is compact implies $F$ is uniformly continuous on $X × X$.

So for $\epsilon > 0$ there exists $\delta > 0$ such that if $d_p((x_1,y_1), (x_2,y_2) ) < \delta$ then $d(F(x_1,y_1), F(x_2,y_2)) < e$.

Here $d_p$ denotes product metrics.

Now let $d(x_1,x_2) < \delta$ then $d_p((x_1,y),(x_2,y)) <\delta$ so $d( F(x_1,y) , F(x_2,y) ) < e$.

Hence, for $d(x_1,x_2) < \delta$ it will be $d(f_y(x_1), f_y(x_2) ) < \epsilon$ for all y in X.

Correct?

If it is duplicate then let me know.