Let $X=C([0,1], \mathbb{R})$, $f \in X$, $\epsilon >0$ and $x_1,...,x_n$ in [0,1]. Consider:
$V_{f,x_1,...,x_n,\epsilon}= \{ g \in X : |f(x_i)-g(x_i)| < \epsilon, i=1,...,n \}$ and $\tau= \{ U \subset X: \forall f \in U, \exists \epsilon>0, \exists x_1, ..., x_n \in [0,1] \ such \ that \ V_{f,x_1,...,x_n,\epsilon} \subset U \}$
I have already shown that $\tau$ is topology and that any sequence in $X$ converges with respect to $\tau$ if and only if it converges pointwise.How do I show that $\tau$ is not metrizable?
This space is not metrizable because it is not first countable. In fact, consider a set of the form $$\left\{V_{0,x_{1k},x_{2k}\ldots,x_{n(k)k},\varepsilon(k)}\,\middle|\,k\in\mathbb N\right\},$$which is a countable set of open sets containing the zero function $0$. Then the set of all $x_{ij}$ is countable and therefore it is distinct from $[0,1]$. Take $y\in[0,1]$ which is not some $x_{ij}$. Then $V_{0,y,1}$ is a neighbourhood of the zero function which does not contain any $V_{0,x_{1k},x_{2k}\ldots,x_{n(k)k},\varepsilon(k)}$.