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?