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.
Yes.
Lemma. Suppose $f\in S$, $x\in(0,1)$, $f$ is continuous at $x$ and $\epsilon>0$. There exists $\delta>0$ such that if $g\in S$ and $|g(x)-f(x)|>\epsilon$ then $||f-g||_1\ge\epsilon\delta/2$.
Proof: Suppose first $g(x)-f(x)>\epsilon$. Choose $\delta>0$ so that if $I=(x,x+\delta)$ then $I\subset(0,1)$ and $$|f(t)-f(x)|<\epsilon/2\quad(t\in I).$$If $t\in I$ then $$g(t)-f(t)\ge g(x)-(f(x)+\epsilon/2)>\epsilon/2,$$so $$||g-f||_1\ge\epsilon\delta/2.$$
If on the other hand $g(x)-f(x)<-\epsilon$ a similar argument works with $I=(x-\delta,x)$.
Corollary. If $f,f_n\in S$ and $||f_n-f||_1\to0$ then $f_n\to f$ almost everywhere on $[0,1]$.
Because a monotone function is continuous almost everywhere.
Exercise. Explain the relevance of the example $$f_n(x)=\begin{cases}
-1+nx,&(0\le x\le 1/n),
\\0,&(1/n\le x\le 1).\end{cases}$$
Best Answer
(1) You only know the convergence $\phi(f_n) \rightarrow \phi(f)$ for bounded linear functionals $\phi$, not for arbitrary linear maps $\phi \in B(C([0,1]), Y)$ (and, as Daniel Fischer points out: What is $Y$?).
Convince yourself that the linear(!), bounded(!) functional $\phi_x(f) := f(x)$ does the job.
(2) Try something like $f_n = n \cdot \chi_{(0, 1/n)}$. Of course, you have to modify the idea to make it a continuous function.