Is the following set compact

Let $S$ be the set $S = \{(e^{-x}\cos (x), e^{-x}\sin(x)) : x\geq 0\} \cup \{(x,0):0\leq x \leq 1\}$.

Is $S$ compact? I know I have to show that if $S$ is closed and bounded, then it is compact. Since the set contains $x=0$ though and $x$ is never less than $0$, is it clearly compact? Or is there something more to show?

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The only limit point of the first set $S_1$ but not in $S_1$ is $(0,0)$. Since for any $p=(u,v)\neq (0,0)$ and $p$ is not an element of the first set $S_1$, let $r^2=u^2+v^2$, we can find sufficient large $x$, s.t $e^{-x}<r$. Then we can find a neighborhood of $p$, which doesn’t intersect $S_1$. Hence $p$ is not a limit point.

Since $(0,0)$ is included in the second set $S_2=\{(x,0):0\leq x \leq 1\}$ and $S_2$ is closed, the union $S=(S_1 \cup \{(0,0)\}) \cup S_1$ should be closed. And it’s bounded since it’s contained in the unit square. Hence it’s compact.