I don't know whether the property I discuss is advanced for the first year student. This example shows that weak$^*$ boundness in $X^*$ imply uniform boundness provided $X$ is complete (see uniform boundness principle). But this is not true in general
Consider the space $c_{00}(\mathbb{N})$ of finitely supported sequences with uniform norm. This is not a complete space, and its completion is $c_0(\mathbb{N})$. Consider family of functionals $\{f_n:n\in\mathbb{N}\}$ defined by equality
$$
f_n :c_{00}(\mathbb{N})\to\mathbb{C}:x\mapsto\sum\limits_{k=0}^nkx(k)
$$
It is easy to check that
Each functional $f_n$ is bounded and $\Vert f_n\Vert=\frac{1}{2}n(n+1)$. Hence
$$
\sup\{f_n:n\in\mathbb{N}\}=+\infty
$$
For each $x\in c_{00}(\mathbb{N})$ the sum $\sum\limits_{k=0}^\infty k|x(k)|$ is finite. Hence we can obtain
$$
\sup\{|f_n(x)|:n\in\mathbb{N}\}\leq\sum\limits_{k=0}^\infty k|x(k)|<\infty
$$
Thus we constructed a family of bounded linear operators (in particular - functionals) which are pointwise bounded but not uniformly bounded. The raison d'etre of such a family is that our space $c_{00}(\mathbb{N})$ is not complete.
On the other hand uniform boundness principle guarantees boundness of any family of functionals in $X^*$ provided the space $X$ is complete.
There are (at least) two somewhat conflicting definitions of a dual space.
In functional analysis, we start with a topological vector space $V$, usually over real or complex numbers, and then the (continuous) dual is the space of all continuous functionals. For normed (in particular, Banach) spaces, a functional is continuous if and only if it is bounded.
In abstract algebra, we deal with vector spaces over arbitrary fields. Frequently, the vector spaces are finite-dimensional, in which case the dual actually coincides with algebraic dual, but for infinite dimensional spaces (over real or complex numbers), the algebraic dual is usually larger by far than the continuous dual*.
More abstractly, the difference of definition is a matter of perspective. In terms of category theory, you can think of the dual of a $K$-vector space $V$ as the space $\operatorname{Hom}(V,K)$, i.e. the set of morphisms from $V$ to the base field $K$ in your category. If the category is that of vector spaces, you obtain the algebraic dual. If the category is that of topological vector spaces, you obtain the continuous dual.
*${}$a pure vector space can be regarded as a discrete topological vector space. In this case, it is fairly easy to see that the continuous dual and the algebraic dual coincide.
Best Answer
By definition $X'$ is a space of bounded linear functionals on $X$. More preciesly $$ X'=\{f\in\mathcal{L}(X,\mathbb{C}):\Vert f\Vert<+\infty\} $$ where $\mathcal{L}(X,\mathbb{C})$ is a linear space of linear functions from $X$ to $\mathbb{C}$ and $$ \Vert f\Vert:=\sup\{|f(x)|:x\in X\quad \Vert x\Vert\leq 1\} $$ In order to prove that $X'$ is complete consider Cauchy sequence $\{f_n:n\in\mathbb{N}\}\subset X'$. Fix $\varepsilon>0$. Since $\{f_n:n\in\mathbb{N}\}$ is a Cauchy sequence there exist $N\in\mathbb{N}$ such that for all $n,m>N$ we have $\Vert f_n-f_m\Vert\leq\varepsilon$. Consider arbitrary $x\in X$, then $$ |f_n(x)-f_m(x)|=|(f_n-f_m)(x)|\leq\Vert f_n-f_m\Vert\Vert x\Vert\leq\varepsilon \Vert x\Vert $$ Thus we see that $\{|f_n(x)|:n\in\mathbb{N}\}\subset\mathbb{C}$ is a Cauchy sequence. Since $\mathbb{C}$ is complete, there exist unique $\lim\limits_{n\to\infty}f_n(x)$. Since $x\in X$ is arbitrary we can define function $$ f(x):=\lim\limits_{n\to\infty}f_n(x) $$ Our aim is to show that $f\in X'$ and $\lim\limits_{n\to\infty}f_n=f$. Let $x_1,x_2\in X$, $\alpha_1,\alpha_2\in\mathbb{C}$ then $$ f(\alpha_1 x_1+ \alpha_2 x_2)= \lim\limits_{n\to\infty}f_n(\alpha_1 x_1+ \alpha_2 x_2)= $$ $$ \lim\limits_{n\to\infty}(\alpha_1 f_n(x_1) + \alpha_2 f_n(x_2))= \alpha_1 \lim\limits_{n\to\infty}f_n(x_1) + \alpha_2 \lim\limits_{n\to\infty}f_n(x_2))= \alpha_1 f(x_1) + \alpha_2 f(x_2) $$ So we conclude $f\in\mathcal{L}(X,\mathbb{C})$. Since $\{f_n:n\in\mathbb{N}\}$ is a Cauchy sequence it is bounded in $X'$, i.e. there exist $C>0$ such that $\sup\{\Vert f_n\Vert:n\in\mathbb{N}\}\leq C$. Hence, for all $x\in X$ we have $$ |f(x)|= |\lim\limits_{n\to\infty}f_n(x)|= \lim\limits_{n\to\infty}|f_n(x)|\leq \limsup\limits_{n\to\infty}\Vert f_n\Vert \Vert x\Vert\leq \Vert x\Vert\sup\{\Vert f_n\Vert:n\in\mathbb{N}\}\leq C\Vert x\Vert $$ Now we see that $\Vert f\Vert\leq C$, but as we proved earlier $f\in\mathcal{L}(X,\mathbb{C})$, so $f\in X'$. Finally recall that for given $\varepsilon>0$ and $x\in X$ there exist $N\in\mathbb{N}$ such that $n,m>N$ implies $$ |f_n(x)-f_m(x)|\leq\varepsilon \Vert x\Vert. $$ Then let's take here a limit when $m\to\infty$. We will get $$ |f_n(x)-f(x)|\leq\varepsilon \Vert x\Vert. $$ Since $x\in X$ is arbitrary we proved that for all $\varepsilon>0$ there exist $N\in\mathbb{N}$ such that $n>N$ implies $$ \Vert f_n-f\Vert= \sup\{|f_n(x)-f(x)|:x\in X,\quad \Vert x\Vert\leq 1\}\leq \varepsilon. $$ This means that $\lim\limits_{n\to\infty} f_n=f$. Since we showed that every Cauchy sequence in $X'$ have a limit, $X'$ is complete.
This proof can be easily generalized up to the following theorem: If $X$ is a normed space and $Y$ is a Banach space, then the linear space of all bounded linear functions from $X$ to $Y$ is complete.