Can every continuous function on complex domain be approximated by polynomials pointwise? Do you know any theorem that will help me with this question:

Let $f$ be any continuous function on complex plane. Show that there is a sequence $(P_n)$ of  polynomials such that $P_n$ converges pointwise to $f$ ($P_n(z) \to f(z)$ for all $z$). 

I tried to use the Runge's approximation  theorem, but I have an arbitrary continuous function, and the theorem requires analyticity on the interior.
I take the rational functions to be with poles only at infinity so they will be polynomials, but I am struggling with the function being only continuous.   
 A: It's impossible to find a sequence of holomorphic functions, never mind polynomials, that converge to $\bar {z}$ pointwise everywhere on the open unit disc $\mathbb {D}.$ Suppose to the contrary we have $f_n(z)\to \bar {z}$ for $z\in \mathbb {D},$ where each $f_n\in H(\mathbb {D}).$ Define $$E_N=\{z\in \mathbb {D}: |f_n(z)-\bar {z}|\le 1,\ \forall n\ge N\}$$ The sets $E_N$ are closed in $\mathbb {D},$ and  $\mathbb {D}=\bigcup E_N$. By Baire, some $E_N$ has nonempty interior, let's call it $U$. We then have $f_N(z), f_{N+1}(z), \dots \to \bar {z},$ boundedly in $U.$ But then from Montel we know this convergence is uniform on compact subsets of $U.$ Hence the limit is holomorphic on $U.$ But $\bar {z}$ is not holomorphic on any nonempty open set, contradiction.
A: In fact, if $f_n$ is a sequence of holomorphic functions converging pointwise to a function $f$ on some domain $\Omega$, a (not as well-known as it should be) theorem by Osgood shows that there must exist an open, dense subset $U \subseteq \Omega$ such that $f$ is holomorphic on $U$ (and that the convergence is in fact locally uniform on $U$).
In particular this for example implies that a nowhere holomorphic function cannot be approximated pointwise by polynomials on any domain.
