I want to know if my proof for the following proposition is correct:
Let $U\subseteq\mathbb{R}\times\mathbb{R}^d$ and $V\subseteq\mathbb{R}^m$ be open sets (usual topology) and let $F:U\times V\rightarrow\mathbb{R}^d$ be a continuous function. Let $(\lambda_n)_{n\in\mathbb{N}} , \lambda_0\in\mathbb{R}^m$ be such that $\lambda_n\rightarrow\lambda_0$.Define $F_n(t,x)=F(t,x,\lambda_n)$ for $n\in\mathbb{N}\cup\{0\}$. Prove that for each $K\subseteq U$ compact we have $F_n|_K\rightarrow F_0|_K$ uniformly
My proof is: Let $\alpha>0$ be such that the closed ball with center $\lambda_0$ and radius $\alpha$, $B_\alpha(\lambda_0)$, is completely in $V$ (possible because $V$ is open), and consider the compact set $K'=K\times B_\alpha (\lambda_0)\subseteq U\times V$. Because $\lambda_n\rightarrow\lambda_0$ eventually $\lambda_n\in B_\alpha (\lambda_0)$ for all $n$ suficiently big, so that we can assume $(\lambda_n)_{n\in\mathbb{N}}\subseteq B_\alpha (\lambda_0)$. Now let $\epsilon>0$. Since $F$ is continuous, it is uniformly continuous in $K'$, hence there exists $\delta>0$ such that for all $(t,x,\lambda),(t',x',\lambda')\in K'$ and $|(t,x,\lambda)-(t',x',\lambda')|<\delta$ then $|F(t,x,\lambda)-F(t',x',\lambda')|<\epsilon$. Since $\lambda_n\rightarrow\lambda_0$, there exists $n_0\in\mathbb{N}$ such that if $n>n_0$ then $|(t,x,\lambda_n)-(t,x,\lambda_0)|<\delta$ for all $(t,x)\in U$. Hence for $n>n_0$ we get: $|F_n(t,x)-F_0(t,x)|<\epsilon$ for all $(t,x)\in U$