Theorem: There is an $r>0$ such that for each $y \in A$, where $A$ is a sequentially compact metric space, $D(y,r) \subset U_i$ for some $U_i$.
Proof: Suppose not. Then for every integer $n$, there is some $y_n$ such that $D(y_n, 1/n)$ is not contained in any $U_i$. By hypothesis, $y_n$ has a convergent subsequence, say, $z_n \rightarrow z \in A$. Since the $U_i$ cover $A$, $z \in U_i$ for some $U_{i_0}$. Choose $\epsilon >0$ such that $D(z,\epsilon) \subset U_{i_0}$ which is possible since $U_{i_0}$ is open. Choose $N$ large enough so that $d(z_N, z) < \epsilon/2$ and $1/N < \epsilon/2$. Then $D(z_N,1/N) \subset U_{i_0}$ a contradiction.
Can someone explain why there exists a contradiction?
We suppose $\forall y\in A$ we have $D(y,r)\subset U_i$ is NOT true. Or in other words, there is some point $y$ where, $\forall r$ we have $D(y,r)\not\subset U_i$ $\forall i$. Now follow the argument. We start by saying, we may index a set of radii around $y$ by observing $y_n$ and the radius $1/n$ which converges to $0$, as we know. But, since $\forall r>0$, $D(y,r)\not\subset U_i$, none of the radius will be inside some open nerighborhood.
But, the sequence $y_n$ is an infinite sequence indexed by the natural numbers. Since $A$ is a sequentially compact metric space, it must have a convergent subsequence (we shall call it $z_n$ and say it converges to $z$ in $A$. In reality, $z=y$ and $z_n=y$ for all of the $n$).
Since $U_i$ cover $A$, $z$ must be in some $U_{i_0}$. This must be true since $z\in A$. Then, for some $\varepsilon >0$, we have $D(z,\varepsilon)\subset U_{i_0}$ since $U_{i_0}$ is an open set (aka it is the union of open disks, so we can choose a radius sufficiently small that will be completely contained in this disk)!
Now, we have a radius $\varepsilon >0$ such that $z\in D(z,\varepsilon)$. Since $z_n$ is a convergent sequence, we may find $N_1\in \mathbb{N}$ such that $\forall n\geq N_1$, $d(z_n,z)<\varepsilon/2$. By the archimedian principle, we may find $N_2\in \mathbb{N}$ such that $1/N_2<\varepsilon/2$. Set, $N=\max\{N_1, N_2\}$. Then we have $D(z_N,1/N)\subset U_{i_0}$ where $U_{i_0}$ is in $A$, and $z_n$ is precisely $y$! This means that our $y$ was in some open set the whole time! But we assumed that it was in NO open set! A contradiction at last.