Question about transitivity convergence Let us define $E^*=\{f:E\rightarrow \mathbb{R} \mid f $ is linear and continuous$\}$, where $E$ is separable space equipped with a norm. Let $A = \{x_1,x_2,\ldots\}$ be a countable dense subset of $E$. Suppose that $(f_n)_{n\in\mathbb{N}}$ is a sequence of functions from $E^*$ such that $$\lim_{n\to\infty}f_n(x_i)=f(x_i)$$for all $i\in\{1,2,\ldots\},$ i.e. $(f_n)$ converges pointwise to $f$ on $A$. My teacher writes then:

Consequently, $f_n(x) \rightarrow f(x) $ $\forall x \in E$.

My question is: why is it true? 
I am trying to demonstrate this as follows:
Choose any $x_i$ close to $x$. Then $f_n(x - x_i) = 0 = f_n(x) -f_n(x_i)$. We have $$\lim_{n\rightarrow \infty}|f_n(x) - f(x_i)| = 0 \iff f_n(x) \rightarrow f(x_i)$$ as $n\rightarrow \infty$. By hypothesis $f_n(x_i) \rightarrow f(x_i)$ whilst by continuity $f(x_i) \rightarrow f(x)$ and  then I will be able to conclude that $f_n(x) \rightarrow f(x) $ $\forall x \in E$. Is it correct?
 A: Your proof is quite strange because it starts from the assumption that $f_n(x-x_i)=0$ for all $n\in\mathbb{N}$. This is not true in general.
I'll additionally assume that $\{f_n:n\in\mathbb{N}\}\subset \mathrm{Ball}_{E^*}(0,C)$ for some $C>0$, i.e. norms of functionals $\{f_n:n\in\mathbb{N}\}$ is uniformly bounded by some constant $C$.
Take arbitrary $x\in E$. Fix $\varepsilon>0$, and find $i\in\mathbb{N}$ such that 
$$
\Vert x- x_i\Vert\leq\frac{\varepsilon}{3\max\{ C,\Vert f\Vert\}}
$$
Since $\lim\limits_{n\to\infty} f_n(x_i)=f(x_i)$, then there exist $N\in\mathbb{N}$ such that $n>N$ implies $|f_n(x_i)-f(x_i)|\leq\varepsilon/3$. Now we use $\varepsilon/3$-trick, i.e. for all $n>N$ we have
$$\eqalign{
|f_n(x)-f(x)|&\leq
|f_n(x)-f_n(x_i)|+|f_n(x_i)-f(x_i)|+|f(x_i)-f(x)|\phantom{A\over B}\cr
&=
|f_n(x-x_i)|+|f_n(x_i)-f(x_i)|+|f(x_i-x)|\phantom{A\over B}\cr
&\leq
\Vert f_n\Vert\Vert x-x_i\Vert+\varepsilon/3+\Vert f\Vert\Vert x-x_i\Vert\phantom{A\over B}\cr
&\leq
C\frac{\varepsilon}{3\max\{ C,\Vert f\Vert\}}+\varepsilon/3+\Vert f\Vert\frac{\varepsilon}{3\max\{ C,\Vert f\Vert\}}\cr
&\leq
\varepsilon/3+\varepsilon/3+\varepsilon/3\phantom{A\over B}\cr
&=\varepsilon}
$$
Thus for all $x\in E$ and $\varepsilon>0$ we found $N\in\mathbb{N}$ such that $n>N$ implies
$|f_n(x)-f(x)|<\varepsilon$. This means that for all $x\in E$ we have $\lim\limits_{n\to\infty} f_n(x)=f(x)$.
P.S.
If we won't require uniform boundedness of norms of $\{f_n:n\in\mathbb{N}\}$ we can construct a counterexample. See Davide's comment.
