Abstract counterexample
Let $E \subset \Omega$ be compact, nowhere dense, and have positive Lebesgue measure (like a fat Cantor set). Then $U = \Omega \setminus E$ is open and dense in $\Omega$. I claim that $L^1(U)$ is weak-* dense in $C_b(\Omega)^*$. This follows from Hahn-Banach, because $L^1(U)$ separates the points of $C_b(\Omega)$: suppose $\phi \in C_b(\Omega)$ is such that $\int \phi f = 0$ for all $f \in L^1(U)$. Take $f = \phi 1_U$; then this says $\int_U \phi^2 = 0$, so $\phi = 0$ almost everywhere on $U$. Since $U$ is open this means $\phi = 0$ everywhere on $U$, and since $U$ is dense we have $\phi = 0$ everywhere on $\Omega$.
In particular, this means you can choose a sequence $f_n \in L^1(U)$ converging weak-* to $f = 1_E$. But clearly $f_n$ does not converge weakly to $f$; take $\phi = 1_E$.
(I cheated a little: $C_b(\Omega)^*$ is not weak-* metrizable, so I cannot necessarily choose a sequence $f_n$ converging weak-* to $f$. You can fix this by choosing a compact regular $\Omega_0 \subset \Omega$ containing $E$ in its interior. Then do everything on $C(\Omega_0)^*$ where the bounded sets are weak-* metrizable.)
Hands-on counterexample
Let's work in one dimension. Let $\Omega = [0,1]$ (or any open set containing $[0,1]$, if you prefer). Let $E \subset [0,1)$ be a fat Cantor set which is closed, nowhere dense, and has positive Lebesgue measure $m(E) > 0$. Set $f = 1_E$. Now for each $n$, partition $[0,1)$ into the intervals $I_{n,j} = [(j-1)/n, j/n)$, $j=1,\dots,n$. Set
$$f_n = \sum_{j=1}^n \frac{m(I_{n,j} \cap E)}{m(I_{n,j} \cap E^c)}1_{I_{n,j} \cap E^c}$$
so that $f_n$ is supported on $E^c \cap [0,1)$ and has the property that $\int_{I_{n,j}} f_n(x)\,dx = \int_{I_{n,j}} f(x)\,dx$ for each $j$.
Clearly we do not have $f_n \to f$ weakly (take $\phi = 1_E$).
But suppose $\phi$ is continuous on $\Omega$, hence uniformly continuous on $[0,1]$. Fix $\epsilon > 0$ and choose $N$ so large that the oscillation of $\phi$ is at most $\epsilon$ on every interval of length at most $1/N$. Then for any $n \ge N$ and any $I = I_{n,j}$
$$\begin{align*}\int_I \phi \cdot (f - f_n) &\le \max_I \phi \int_I f - \min_I
\phi \int_I f_n \\
&= (\max_I \phi - \min_I \phi) \int_I f \\
&\le \epsilon \int_I f. \end{align*}$$
Summing over $j$ we have $\int \phi \cdot (f_n - f) \le \epsilon \int f = \epsilon m(E)$. A similar argument gives $\int \phi \cdot (f_n -f) \ge -\epsilon m(E)$. So we have $\left|\int \phi \cdot (f_n - f) \right| \le \epsilon m(E)$, and this shows that $\int \phi f_n \to \int \phi f$. So $f_n \to f$ in your weak-* sense.
For the first part, your argument works if you define $\mu(B)=\int_B \lvert \psi\rvert d\lambda$, where $\lambda$ denotes the Lebesgue measure.
Alternatively, you can approximate in $\mathbb L^1$ by a linear combination of indicator functions of Borel sets of finite measure to get the result.
- Denote $$\left\langle u,v \right\rangle := \int_{\mathbb R}u(x)v(x) dx$$
and $g_n(x):=\operatorname{sgn}\left(f_n(x)\right)\mathbf{1}_{\mathbb R\setminus I_n}(x)\lvert \psi(x)\rvert $, where $\operatorname{sgn}(y)=1$ if $y>0$, $\operatorname{sgn}(y)=-1$ if $y<0$ and $\operatorname{sgn}(0)=0$.
Then $$\int_{\mathbb R \setminus I_n} |f_n(x) \phi(x)| dx=\left\langle f_n,g_n \right\rangle $$ and since $g_n\to 0$ strongly in $\mathbb L^2$ (by the first question), we get the wanted result.
Best Answer
This is false. Consider, for example, $f_n=n^{3/4}\chi_{(0,1/n)}$ on $D=[0,1]$. Then $f_n\to 0$ in $L^1$, but the sequence does not converge weakly in $L^2$ (it would be bounded in norm if it did).
The problem with your argument is that you are implicitly assuming that $\|f_n\|_2$ is bounded; otherwise, there's no obvious way to carry out the approximation argument going from simple functions to general functions.