Proof of the Banach–Alaoglu theorem The Banach–Alaoglu theorem states that the closed unit ball of $B'$ (where $B'$ is the dual to a Banach space $B$ over a field) is compact in the weak* topology. I'm having trouble trying to prove the theorem.
I first considered the field to be $\mathbb{C}$, and for each $x$ in $B$, we let $U_{x}$ be the closed ball of radius $||x||_{b}$ in $\mathbb{C}$. Then, we can take $U = \Pi_{x\in B} U_{x}$ which will be compact by the Tychonov Product Theorem in the product topology (so, an element of U would be $(z_{x})_{x \in B}$).
Then I saw a proof, in where we can define for $x_{1}, x_{2} \in B$ and $t_{1}, t_{2} \in \mathbb{C}$:
$$E_{x_{1}, x_{2}, t_{1}, t_{2}}= \{(z_{x})|z_{t_{1}x_{1}+t_{2}x_{2}}=t_{1}z_{x_{1}}+t_{2}z_{x_{2}}\}$$
Then, it is stated that the above condition defines a closed subset of $$U_{t_{1}x_{1}+t_{2}x_{2}} \times U_{x_{1}} \times U_{x_{2}}$$ (I don't see the reason why) and hence $E_{x_{1}, x_{2}, t_{1}, t_{2}} \subset D$ is closed as well (which I don't understand why as well). Now, define:
$$E=\displaystyle\bigcap_{x_{1},x_{2}\in B, t_{1},t_{2}\in \mathbb{C}} E_{x_{1},x_{2},t_{1},t_{2}}$$
Which is clearly closed. Then, it is stated that there is a bijection between the closed unit ball of $B'$ and $E$ given by $u \mapsto (z_{x})$, where $z_{x}=u(x)$, $\forall x \in B$ which shows $B'$ is compact in the weak star topology. However, how would this be a continuous bijection? (if this is the case, how does this prove the compactness of $B'$?) 
Thanks for your help.
 A: Not sure what $E$ is doing there, but here is how the proof usually goes :
The map
$$
\tau : f \mapsto (f(x))_{x\in X}
$$
defines a continuous injection from the unit ball of $B'$ to $U$.
Let $A$ denote the image of $\tau$, so it suffices to prove that $A$ is closed in $U$. Suppose $y \in U$ such that $\tau(f_{\alpha}) \to y$. Then, for every $x \in B$ with $\|x\| \leq 1$, $f(x) := \lim f_{\alpha}(x)$ exists. Now extend $f$ to a linear functional on $X$ by scaling.
Now note that if $\|x\|\leq 1$, then $|f(x)| \leq 1$, so $f$ is continuous as well, and $\tau(f) = \lim \tau(f_{\alpha})$ in $A$.
Hence, $A$ is closed, and this completes the proof.
A: Here is an answer to the specific question that was asked (I myself am coincidentally working through the same proof). Allow me to add to your notation by defining the projections $\pi_x : U \mapsto U_x$ by $\pi_x(z) = z_x \in U_x$, which are continuous in the product topology. Now consider the set in question
$$E_{x_1,x_2,t_1,t_2} = \{ z \in U : \pi_{t_1x_1+t_2x_2}(z) = t_1\pi_{x_1}(z) + t_2\pi_{x_2}(z)\} $$
Our goal is to see that the above set is closed. We can rewrite it as
$$ E_{x_1,x_2,t_1,t_2} = \{z \in U : (\pi_{t_1x_2+t_2x_2} - t_1\pi_{x_1} - t_2\pi_{x_2})(z)=0\}$$
and it is not difficult to see that the linear combination of projections above is continuous. Thus $E_{x_1,x_2,t_1,t_2}$ is the inverse image of a closed set (the singleton $\{0\}$) under a continuous mapping, which means is it is closed in $U$. This means that the arbitrary intersection of these $E$'s over all $x_1,x_2,t_1,t_2$ is also closed. Since $U$ is compact (Tychnoff's Theorem), and $E\subset U$, $E$ is compact. I will leave it to you to come up with the homeomorphism from 
$$ E=\bigcap_{x_1,x_2\in X, t_1,t_2 \in \mathbb{C}} E_{x_1,x_2,t_1,t_2}$$
to the closed unit ball of $B'$ (at least I hope you can convince yourself that they're bijective).
Edit: In fact, I believe you only need to exhibit a continuous bijection from $E$ to the closed unit ball in $B'$ - this follows from a lemma that states that a continuous bijection from a compact space to a Hausdorff (T2) space is automatically a homeomorphism. The map identification should not be hard to come up with!
