Let $f$ be an integrable function in $[a,b]$. Define, for $n \in \mathbb{N}$,
$$I_{n}(f) := \sum_{j=0}^{n} w_{j,n}f(x_{j,n})$$ to numerically approximate $\int_{a}^{b} f(x)dx$. (In the context of Weierstrass theorem) prove that if:
$I_{n}(f) \rightarrow \int_{a}^{b} f(x)dx$ for all function in some dense family of $C[a,b]$ and $\sup_{n} \sum_{j=0}^{n} w_{j,n}< \infty $.
Then $I_{n}(f) \rightarrow \int_{a}^{b} f(x)dx$ for all $f \in C[a,b]$.
I think I have to use Lagrange interpolation to relate some concepts for the proof but I don't know how to use Weierstrass theorem.
This is a hint is rather than a full solution. This is a problem where you must assert control over the term $I(f) - I_n(f)$ where $I(f) = \int_a^b f(x) dx$. To that end, you must introduce terms over which you have control. By assumption, there is a sequence of functions $\{f_m\}_{m=1}^\infty$ such that $f_m \rightarrow f$ uniformly on $C([a,b])$ and such that $I_n(f_m) \rightarrow I(f_m)$ for $n \rightarrow \infty$ and each fixed value of $m$. This suggest to me that we should write $$ I(f) - I_n(f) = \big[ I(f) - I(f_m) \big] + \big[I(f_m) - I_n(f_m) \big]+ \big[I_n(f_m) - I_n(f)\big].$$ Now try to assert control over each of the three terms in the square brackets. You will need to use your assumption about the weights $W_{j,n}$. Check to see if you are allowed to assume that the $w_{j,n}$ are non-negative.