Define a subset $X$ of $\mathbb{R}^n$ to have property C if every sequence with exactly one accumulation point in $X$ converges in $X$. (Recall that $x$ is an accumulation point of a sequence $(x_n)$ if every neighborhood of $x$ contains infinitely many $x_n$.)
Show that any subset $X$ of $\mathbb{R}^n$ satisfying property C is compact.
Since $\mathbb{R}^n$ is Euclidean, compactness $\iff$ closed and bounded. To show $X$ is closed, we can take any limit point $x$ of $X$. By definition of a limit point, we can find a sequence in $X$ converging to $x$. Then $x$ is the unique accumulation point of the sequence and hence by hypothesis $x\in X$
How can I show that $X$ is bounded?
Alternatively, I could show that $X$ is sequentially compact, but I don't know how to proceed.
You need to show that $C$ is closed and bounded.
If it is unbounded, then let $\{x_n\}\subset C$ be an unbounded sequence, such that $\|x_n\|>n$. Set $$\cases{y_{2n}=x_1\\y_{2n+1}=x_n}$$ Then because of the requirement $\|x_n\|>n$, $x_1$ is the only accumulation point of $y$, which still doesn't converge.
If $C$ is not closed, then let $\{x_n\}\subset C$ be a sequence with limit $x\not\in C$. Then, again, set $y$ as before. $x_1$ is the only accumulation point of $y$, yet $y$ doesn't converge.