A slicker proof (and more importantly one that generalizes) than the one in the wikipedia article is to observe that $X_n \Longrightarrow X$ if and only if for all bounded continuous functions $f$ we have $E f(X_n) \to E f(X)$. If you have convergence in probability then you can apply the dominated convergence theorem (recalling that $f$ is bounded and that for continuous functions $X_n \to X$ in probability implies $f(X_n) \to f(X)$ in probability) to conclude that $E |f(X_n) - f(X)| \to 0$, which implies the result.
The function $g=\mathbb{1}_{(0,1)}$, which takes values $0$ outside $(0,1)$ and $1$ in $(0,1)$ will work for your purposes.
Notice that the points of discontinuities of $g$ is $\{0,1\}$, and $\mathbb{P}[X\in\{0,1\}]=1$. Furthermore, $Z_n=g(X_n)=1$ and $Z=g(X)=0$.
In terms of measures (no random variables involved) define the sequence of probability measures $\mu_n$ on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$ as
$$\mu_n=\frac12\delta_{\frac1n}+\frac12\delta_{1-\frac1n}$$
For any $f\in\mathcal{C}_b(\mathbb{R})$ (here $\mathcal{C}_b(\mathbb{R})$ is the space of real valued bounded continuous functions on $\mathbb{R}$)
$$\int f\,d\mu_n=\frac12 f(1/n)+\frac12 f(1-1/n)\xrightarrow{n\rightarrow\infty}\frac12(f(0)+f(1))$$
Consequently $\mu_n\stackrel{n\rightarrow\infty}{\Longrightarrow} \frac12(\delta_0+\delta_1)=:\mu$.
With $g(x)=\mathbb{1}_{(0,1)}(x)$ on $\mathbb{R}$, and $f\in\mathcal{C}_b(\mathbb{R})$
$$\int f\circ g\,d\mu_n=\frac12\big(f(g(1/n))+f(g(1-1/n))\big)=f(1),\qquad n\geq2$$
Thus, $\mu_n\circ g^{-1}\Longrightarrow\delta_1$ however, $\mu\circ g^{-1}=\delta_0$. As I mentioned before, that set the points of discontinuity of $g$ is given by $D=\{0,1\}$, and $\mu(D)=1$.
Best Answer
I suspect he meant to say $F_n(x_0) \not\to F(x_0)$. Observe that $F_n(x_0) = 1$ for every $n$ but $F(x_0) = 0$.
However, as you say, intuitively we really want for $F_n$ to "converge" to $F$. So asking for $F_n(x) \to F(x)$ for every $x$ is too strong, and thus we weaken the definition by only requiring it to hold at points $x$ where $F$ is continuous. This definition turns out to be a better fit for what we are trying to describe.