continuous linear functional on a reflexive Banach space attains its norm

How does one prove that if a $X$ is a Banach space and $x^*$, a continuous linear functional from $X$ to the underlying field, then $x^*$ attains its norm for some $x$ in $X$ and $\Vert x\Vert = 1$?

My teacher gave us a hint that we should use the statement that if $X$ is a reflexive Banach space, the unit ball is weak sequentially compact, but I am not sure as to how to construct a sequence in this ball which does not converge.

Thank you.

We can use a corollary of Hahn-Banach theorem, applied to the dual space $X'$ of $X$. We have $$\lVert x'\rVert=\max_{y\in X'',\lVert y\rVert=1}|y(x')|$$ (the maximum is reached for a $y_0$ that can be constructed thanks to Hahn-Banach theorem). For this $y_0\in X''$, since $X$ is reflexive we can find $u\in X$ such that $J(u)=y_0$, where $J\colon X\to X''$ is the canonical embedding. Hence by definition $J(u)(x')=x'(u)=y(x')$ and $u$ (or $-u$) is such that $|x(u)|=\sup_{\lVert v\rVert=1}|x(v)|$.

Note that it doesn't follow the hint given by your teacher. Note that the converse is true (if each linear continuous functional attains its norm, the Banach space is reflexive). It's a difficult result, from James I think.

Suppose that $\varphi \neq 0$. Let $x_n$ be a sequence in the unit sphere such that $|\varphi(x_n)| \geq \lVert \varphi\| - \frac{1}{n}$ which exists by the definition of the operator norm. Choose a subsequence such that $x_{n_k} \to x$ weakly with $\|x\| \leq 1$ by weak sequential compactness of the unit ball. Deduce that $|\varphi(x)| = \|\varphi\|$. Then argue that $\|x\| = 1$.