The key is the following result:
If $X$ is a normed space, and $\{x_n\}\subset X$ a sequence which converges weakly to $X$, then $\sup_{n\in\Bbb N}\lVert x_n\rVert<\infty$.
It's a consequence of Baire's categories theorem applied to $X^*$, the topological dual of $X$.
We show that an unbounded set cannot be weakly compact. If $S$ is not bounded, let $\{x_n\}\subset S$ such that $\lVert x_n\rVert\geqslant n$. If $\{x_{n_k}\}$ is a subsequence of $\{x_n\}$, then for each $k$, $\lVert x_{n_k}\rVert\geqslant n_k$. In particular, $\{x_{n_k}\}$ is not bounded hence cannot be weakly convergent.
Let me give first the definitions:
D A metric space $M$ (or subset of a metric space) is said to be precompact if every sequence in $M$ contains a Cauchy (or fundamental) subsequence.
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D Let $S$ be any subset of $\Bbb R^n$. Given $\epsilon >0$, we say that $N$ is an $\epsilon$-net for $S$ if the set of open balls
$$B_\epsilon(N)=\{B(x,\epsilon):x\in N\}$$
covers $S$. That is, the set of open balls of radius $\epsilon$ centered at the points of $N$ cover $S$.
Now, a theorem by Hausdorff
T Let $P$ be a metric space, $M$ a subset of $P$. Then $P$ is precompact if and only if, given $\epsilon >0$, $P$ contains a finite $\epsilon$-net for $M$.
P Suppose $M$ is precompact. Then given $\epsilon >0$, choose $x_1\in M$. If the point $x_1\in M$ is such that $d(x,x_1)<\epsilon$ for each $x\in M$; $\{x_1\}$ is itself an $\epsilon$-net, and we're done. If there exists points with $d(x,x_1)>\epsilon$, choose as $x_2$ one of these points. If now for every $x\in M$ we have either $d(x,x_1)<\epsilon$ or $d(x,x_1)<\epsilon$ , we're done. Else, continue the process, noting that the distance from each new $x_n$ to the $x_i\; ;1\leq i\leq n-1$ exceeds $\epsilon$. Thus, if the construction fails to terminate after a finite number of steps, we would obtain a sequence $\{x_n\}_{n\in \Bbb N }$ which contains no Cauchy subsequence, contrary to the hypothesis that $M$ is precompact. Thus, the construction must terminate, providing us with a finite $\epsilon$-net, $\{x_1,\dots,x_{\ell}\}$.
Conversely, suppose that for any $\epsilon>0$, $P$ contains a finite $\epsilon$-net for $M$, and let $A$ be any infinite subset of $M$, in particular, a sequence containing infinitely many distinct points of $M$. We can select a Cauchy subsequence as follows. Let $x_0\in A$ be any point, then take $\epsilon =1$ in the condition and cover $A$ with a finite number of balls of radius $1$. One of these balls, say $B_1$, must contain an infinite $A_1$ of $A$. Choose $x_1\in A_1\setminus \{x_0\}$. Choosing $\epsilon = 1/2$; we cover $A_1$ with a finite number of balls of radius $1/2$. One of them, say $B_2$ must contain an infinite subset $A_2$ of $A_1$. Choose $x_2\in A_2\setminus \{x_0,x_1\}$. Continuing this process, we obtain a sequence $A_0\supseteq A_1\supseteq A_2\cdots$, where each $A_n$ is contained in a ball of radius $1/n$ and a sequence of distinct points $x_0,x_1,x_2,\dots$ with $x_n\in A_n$. This sequence is Cauchy, for if $m<n$, $$B_m\supseteq A_m\supseteq A_n$$ which means $$d(x_n,x_m)<\frac 2 m$$Since $2/m\to 0$, we're done. It follows $M$ is precompact.
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Look here for
T Let $S$ be bounded in $\Bbb R^n$. Then $S$ is precompact.
Now, since $\Bbb R^n$ is complete, any Cauchy (sub)sequence will converge. The assumption that the set is closed means the limit $\ell$ will be in such set. Thus, every bounded closed subset of $\Bbb R^n$ contains a convergent subsequence, that is, it is sequentially compact.
NOTE Observe the similarity of the $\implies$ direction of Hausdorff's criterion with the proof of Weiertrass-Bolzano that Ragib suggests.
Best Answer
Take $A=\{z\in\mathbb{C}\,|\,\lvert z\rvert\geqslant1\}$, for instance. It is closed. However, it is not sequentially compact because, for instance, the sequence $1,2,3,\ldots$ has no convergente subsequence. All you can deduce about sequences of elements of $A$ from the fact that $A$ is closed is that if one such sequence converges, then its limit belongs to $A$.