$A \in \mathcal{L}(X,Y) \implies A^* \in \mathcal{L}(Y^*_{w^*}, X^*_{w^*})$ Exercise :

Let $X,Y$ be Banach spaces and $A \in \mathcal{L}(X,Y)$. Show that $ A^* \in \mathcal{L}(Y^*_{w^*}, X^*_{w^*})$.

Attempt :
The linearity is trivial. Τo show that $A^*$ is $w^*$ to $w^*$ continuous :
Let $g_n \xrightarrow{w^*} g \in Y^*$. For any $x \in X$ it is $A(x) \in Y$ which implies :
$$\langle A^*(g_n), x\rangle = \langle g_n, A(x)\rangle \xrightarrow{w^*} \langle g,A(x)\rangle=\langle A^*(g),x\rangle$$
Thus, it is $A^*(g_n) \xrightarrow{w^*} A^*(g)$, which means that $A^*$ is $w^*$ to $w^*$ continuous.
Q : About boundedness, how does one imply that as we're working over metric spaces equipped with the weak* topology ? Is it enough to show that they are w* to w* continuous since the adjoint operator of a bounded operator is bounded anyway ? Any inputs will be appreciated.
 A: You can use any notation you want as long as you define it. Without explicitly defining $\mathcal{L}(\cdot, \cdot)$ is often used to denote bounded linear operators from a Banach space to another. $C(M,N)$ is a notation that is widespread to denote continuous functions from $M$ to $N$. One could use $C(Y_{w^*}^*,X_{w^*}^*)$. There are some books that use $C$ for closed operators, or compact. The best option is always to define explicitly the notation.
If you want to prove that $A^*:Y_{w^*}^*\to X_{w^*}^*$ is continuous when the weak-$*$ topologies of $Y^*$ and $X^*$ are used, then you can do similar to what you did, except that you should use nets instead of sequences.
If $g_{\alpha}\in Y^*$ is a net converging to $g\in Y^*$ in the weak-$*$ topology, then $\forall y\in Y$ $g_\alpha(y)\to g(y)$ as a net in $\mathbb{C}$.
Then $A^{*}(g_\alpha)\in X^*$ and for all $x\in X$ 
$$A^{*}(g_\alpha)(x)=g_\alpha(A(x))\to g(A(x))=A^*(g)(x)$$
Therefore $A^*(g_\alpha)$ converges to $A^*(g)$ in the weak-$*$ topology of $X^*$.
This is the same argument you did, except with the clarification that we are using nets instead of just sequences, and using the functional notation instead of the bracket notation, but that is is only because it takes me more time to type the $\langle$ and $\rangle$.
The need for nets instead of sequences is due to the weak-$*$ topology not being necessarily first-countable. This can happen, for example, if $X$ and/or $Y$ have uncountable dimension.

Proof that continuous implies bounded
This is general for any continuous linear transformation between locally convex topological vector spaces. 
Since $A^*$ is continuous, for every $x\in X$ and $e>0$, there is a finite number of $y_1,...,y_n\in Y$ and $e_1,...,e_n>0$ such that $A^*(\bigcup_{i=1}^n\{y^*:\ |y^*(y_i)|<e_i\})\subset \{x^*:\ |x^*(x)|<e\}$.
This means that $\forall i=1,...,n, |y^*(y_i)|<\min_{i\in\{1,..,n\}}(e_i)$ implies $|A^*(y^*)(x)|<e$.
Therefore, scaling the $y^*$ we get $$|A^*(y^*)(x)|\leq \frac{e}{\min_{i\in\{1,..,n\}}e_i}\max_{i\in\{1,..,n\}}|y^*(y_i)|$$
