Problem: Suppose you have a sequence $(x_n)\subset X$, where $X$ is a Banach space. Further suppose that for all $\phi\in X^*,\,(\phi(x_n))$ is a bounded sequence in the complex numbers. Show, using the Uniform Boundedness Principle (UBP), that $(x_n)$ is bounded in $X$.
My attempt: We recall UBP:
"For a Banach Space $X$ and a Normed Vector space $Y$, suppose that for $\alpha\in A$, $T_\alpha\in B(X,Y)$ and that $\sup_\alpha\|T_\alpha x\|<\infty,\,\forall x\in X$. Then, $\exists M:\|T_\alpha x\|\le M\|x\|,\,\forall x\in X, \alpha\in A.$"
The last sentence meaning that we can deduce that $\|T_\alpha\|\le M.$
Now, we know that for all $\phi\in X^*$, $(\phi(x_n))$ is bounded in $\mathbb C$. That is to say $\forall \phi\in X^*,\exists M_1\in\mathbb R:|\phi(x_n)|\le M_1,\forall n\in\mathbb N$.
But since this is the case for all continuous linear functionals in $X^*$, taking the $\sup$ over $\phi\in X^*$ will make sense and be finite. And so, $\sup_{\phi\in X^*}|\phi(x_n)|\lt\infty,\forall x\in X.$
This then allows us to apply UBP to deduce that, $|\phi(x_n)|\lt L\|x_n\|,\,\forall\phi\in X^*,x\in X$. And this in turn means that we must have $\|\phi\|\le L.$ But this can only be finite if $\|x_n\|$ is finite for all choices of sequences in $X$. That is to say $\exists M_2\in \mathbb R:\|x_n\|\le M_2$ which precisely means that $(x_n)$ is bounded in $X$.
Is this correct? Is my reasoning sound, and have I employed UBP in the correct manner?
You're not apllying the UBP correctly. While it is true that $\sup_n|\phi(x_n)|<\infty$ for all $\phi\in X^*$ and $n\in\mathbb N$, to say that $|\phi(x_n)|\leq L\|x_n\|$ for all $\phi\in X^*$ is ridiculous (unless $x_n=0$ for all $n$).
As a hint to push you in the right direction, regard the maps $\phi\mapsto \phi(x_n)$ as a sequence of maps in $X^*$, and apply the UBP to obtain something of the form $|\phi(x_n)|\leq M\|\phi\|$ for all $\phi\in X^*$.