Proof of Banach Alaoglu without nets 
(Folland's page 169) Let $X$ be a normed vector space. Let $B^*:= \{ f \in X^* \,: \, ||f|| \le 1 \}$  be the unit ball in $V^*$ under the operator norm.  $B^*$ is compact in $X^*$ in the weak$^*$ topology. 

In his proof, he stated that 

We may identify $B^*$ with $D:= \prod D_x$, where $D_x:= \{ z \in \mathbb{C} \, : \, |z| \le ||x|| \}$. We would like to show that $B^*$ is closed in $D$.

Here is my attemt of proof :

  
*
  
*Define 
  $$ \phi_{x,y,\lambda}: D \rightarrow \mathbb{F}, f \mapsto f(x+\lambda y)-f(x) -\lambda f(y) $$
  
*$\phi_{x,y,\lambda}$  is continuous as it is the composition of the maps (supposing, $x,y,x+\lambda y$ are distinct)
  $$ D \rightarrow \mathbb{F}_x \times \mathbb{F}_y \times \mathbb{F}_{x+\lambda y } \rightarrow \mathbb{F}$$
  The first map $f \mapsto (f(x), f(y),f(x+\lambda y))$ is continuous by definition universal property/definition of product topology. Continuity of last map follows from property of topological vector space. 
  
*$B^* = \bigcap_{x,y \in X, \lambda \in \mathbb{F}} \ker \phi_{x,y,\lambda} $ is closed in inherited product topology. 

Is this proof correct? 
 A: For any $x\in X$, let 
\begin{equation}
 D_{x}=\lbrace z\in ℂ:\vert z\vert‎\leqslant‎\Vert x\Vert \rbrace\\
\end{equation}
and $D=\prod_ {x\in X} ⁢D_{x}$. Since $D_{x}$ is a compact subset of ℂ, D is compact in product topology by Tychonoff theorem.
We prove the theorem by finding a homeomorphism that maps the closed unit ball 
$
B_{{X^\ast}}
$ of $X^{*}$ onto a closed subset of D. Define
\begin{equation}
 \varphi_{x}:B_{X^{\ast}} \longrightarrow‎‎ D_{x} , \varphi x⁢(f)=f⁢(x) , \varphi:B_{X^{\ast}}\longrightarrow D\\
\end{equation}
 by
\begin{equation}
 \varphi=\prod_{x\in X⁢} \varphi{x},
\end{equation}
 so that 
\begin{equation}
\varphi⁢\left(f \right)=\left(f⁢(x)\right) x\in X.\\
\end{equation}
Obviously, $\varphi $ is one to one, and a net $\left(f_{α}\right)\in B_{X^{*}}$ converges to f in weak$^{*}$ topology of $X^{*}$ if  $\varphi \left(f_{α}\right)$ converges to $\varphi⁢(f)$ in product topology, therefore $ \varphi $ is continuous and so is its inverse
\begin{equation}
 \varphi^{-1}:\varphi\left(B_{X^{*}}\right) \longrightarrow  B_{X^{*}}\\
\end{equation}
It remains to show that
 $\varphi \left(B_{X*}\right)$  is closed. If
\begin{equation}
\varphi \left(f_{α}\right)
\end{equation}
 is a net in $\varphi \left(B_{X*}\right)$, converging to a point $ d=(d_{x}) $, $ x\in X\in D$, we can define a function 
\begin{equation}
f: X \longrightarrow C   , f⁢(x)=d_{x}.\\
\end{equation}
 As $\lim_{α}\varphi⁢\left(f{α⁢(x)}\right)=d_{x}$ for all $ x\in X$ by definition of  weak$^{*}$ convergence, one can easily see that f is a linear functional in $ B_{X*}$ and that $\varphi(f)=d$. This shows that d is actually in $\varphi(B_{X^{*}})$ and finishes the proof. 
‎‎
A: Yes, your proof is correct and is in fact the usual proof that is given that doesn't use nets.
