I need to show that a closed, bounded set having exactly one accumulation point has the covering property.
A set has the covering property if any open cover of it has a finite subcover.
Since the set is bounded, isn't it simply contained in $(V, U)$ for some $V$, $U$. Thus it has the covering property? I'm definitely missing something... Not sure where the accumulation point comes in.
Let's call the set (which you implicitly assumed lying in a Euclidean space) $E$, and its sole accumulation point $a$. Since $E$ is closed, it contains $a$.
Suppose $\{U_\alpha:\alpha\in I\}$ is an open cover of $E$. We are to prove that it's possible to cover $E$ using only finitely many of the sets $U_\alpha$.
The point $a$, being in $E$, is covered by something. Formally: there exists $\alpha_0\in I$ such that $a\in U_{\alpha_0}$. It is reasonable to use $U_{\alpha_0}$ for our finite cover. It remains to cover the rest of $E$, namely the set $E\setminus U_{\alpha_0}$.
But here's the key point: $E\setminus U_{\alpha_0}$ is finite, because it's bounded and has no accumulation points. So we enumerate its points $x_1,\dots,x_n$ and for each $i=1,\dots,n$ pick $U_{\alpha_i}$ that contains $x_i$.
End result: $\{U_{\alpha_0},U_{\alpha_1},\dots,U_{\alpha_n}\}$ is the required finite cover.