Here is a proof of Alaoglu's theorem which I find nice (it might actually be the standard proof):
If $X$ is a normed vector space, the closed unit ball $B^*=\{f\in X^* : \|f\|\leq 1\}$ in $X^*$ is compact in the weak* topology (this is taken from Folland's Real Analysis, Theorem 5.18).
Proof
Given a point $x\in X$, consider the set
$$D_x=\{z\in\mathbb{C}:|z|\leq\|x\|\}.$$
Note that $D_x$ is compact for all $x$. Let $D=\prod_{x\in X}D_x$. By Tychonoff's theorem, $D$ is also compact. Now we take a new point of view. Namely, we note that $D$ coincides with the set of all complex-valued functions $\phi$ on $X$ (not necessarily continuous) such that $|\phi(x)|\leq\|x\|$ for all $x\in X$, and $B^*$ is now a subset of $D$ consisting of those functions which are linear. By definition, the weak* topology on $B^*$ is the topology of pointwise convergence, which is exactly the topology $B^*$ inherits from $D$. Therefore, to conclude that $B^*$ is compact (in the weak* topology), it suffices to show that it is a closed subset of $D$.
Let $\{f_\alpha\}$ be a net in $B^*$ that converges to $f\in D$. Then, for all $x,y\in X$ and $a,b\in\mathbb{C}$, we have
$$f(ax+by)=\lim f_\alpha(ax+by)=\lim (af_\alpha(x)+bf_\alpha(y))=af(x)+bf(y),$$
that is, a limit of linear maps is linear. Therefore, we conclude that $f\in B^*$, and so $B^*$ is closed in $D$.
The reason I like this proof is that you have to consider an object in two different ways, and it is the interplay between the different viewpoints that allows you to prove the result. When I first learned functional analysis (and I am still learning), the different topologies kept confusing me, and proofs like this one allowed me to understand these concepts.
Best Answer
It has been a while since I studied it, but I've used this as a course text to a course I couldn't attend the lectures of, and it wasn't exceptionally hard to pass. Your comments give no immediate reason to suspect it would be a bad choice; but be prepared for a quite substantial amount of proofs "left to the reader".