Let $f: \Bbb R \times [0,1] \to \Bbb R$ be a continuous function and $(x_n)$ be a sequence in $\Bbb R$ that converges to $x$ in $\Bbb R$. Then show that the sequence of functions $g_n$ defined by,$$g_n(y)=f(x_n,y) \;\;\;\; 0 \le y \le 1$$
$$g(y)=f(x,y) \;\;\;\;\;\;\;\; 0 \le y \le 1$$
is uniformly convergent to $g$.
My attempt :
Since the set $\{x_n\}_{n \in \Bbb N} \times [0,1]$ is closed and bounded, it is compact. Hence $f$ is uniformly continuous on $\{x_n\}_{n \in \Bbb N} \times [0,1]$.
Consider the sequence $((x_n,y))_{n \in \Bbb N}$ for an arbitrary $y \in [0,1]$. It converges to the point $(x,y)$. Therefore, for a given $\epsilon \gt 0$, $\exists \delta(\epsilon) \gt 0$ and $n_0(\delta(\epsilon))=n_0(\epsilon) \in \Bbb N$ such that for all $y \in \Bbb [0,1]$ and $\;\forall n \ge n_0(\epsilon)$,
$||(x_n,y)-(x,y)|| \lt \delta(\epsilon) \Rightarrow |f(x_n,y)-f(x,y)| \lt \epsilon \Rightarrow |g_n(y)-g(y)| \lt \epsilon.$
Note that $n_0(\epsilon)$ and $\delta(\epsilon)$ depend only on $\epsilon$.
(I have tried to combine sequential continuity and uniform continuity of $f$ in the above proof)
Have I done it right? Are there any loopholes in this proof? Thanks.