Weak Convergence If and Only If (Pointwise) Convergence of Characteristic Function 
This is actually a theorem from lecture notes, with the corresponding proof. Unfortunately, it doesn't prove the last bit, or mention it at all (!), and I have a question about the penultimate bit. This is the definition of weak convergence being used:

Let $\mu$ be a Borel probability measure on $\Bbb R^d$ and let $(\mu_n)$ be a sequence of Borel probability measures on $\Bbb R^d$.
We say that $(\mu_n)$ converges weakly to $\mu$, written $\mu_n \Rightarrow \mu$, if $\mu_n(f) \rightarrow \mu(f)$ as $n \rightarrow \infty$ for all bounded, continuous functions $f:\Bbb R^d \rightarrow \Bbb R$.
We say that the sequence $(X_n)$ of random variables on $\Bbb R^d$ converges weakly to $X$ if $\mu_{X_n} \Rightarrow \mu_X$.

Firstly, consider the penultimate claim (If a sequence...). What I don't understand is why this example is not a counter-example to the claim. ($\phi$ is the characteristic function.)
Secondly, consider the final claim (Conversely, if...). No proof was given in the lecture notes. I think I have a solution, but I'm not sure; if someone could look over my answer (given as an answer below), then I'd be most appreciative. Thanks!
 A: Suppose $X_n \Rightarrow X$, ie $\mu_{X_n} \Rightarrow \mu_X$. We desire to show that
$\phi_{X_n}(\xi) \rightarrow \phi_{X_n}$ as $n \to \infty$ $\forall \xi \in \Bbb R^d.$ Write $\mu_n = \mu_{X_n}$ and $\mu = \mu_X$.
Since, by definition, $\mu_n \Rightarrow \mu$, in particular we have that, for each (fixed) $\xi \in \Bbb R^d$, $\mu_n(e^{i x \cdot \xi}) \rightarrow \mu(e^{i x \cdot \xi})$. Thus,
$$ \begin{align} |\phi_{X_n} - \phi_X| & = |\hat \mu_{X_n}(\xi) - \hat \mu_X(\xi) | \\
& = | \int_{\Bbb R^d} e^{i x \cdot \xi} \, d\mu_{X_n}(x) - \int_{\Bbb R^d} e^{i x \cdot \xi} \, d\mu_{X_n}(x) | \\
& = | \mu_n(e^{i x \cdot \xi}) - \mu(e^{i x \cdot \xi}) | \\
& \rightarrow 0.
\end{align}$$
as $n \to \infty$. (No need for the dominated convergence theorem.)
Also, this is for fixed $\xi$, so I have shown pointwise, not uniform convergence.
A: Note the definition of the characteristic function:
$$\varphi_X(x) = \mathbb{E}(e^{i\langle x, X \rangle}).$$
The weak convergence of $(X_n)$ towards X means that
$$\lim_{n\to\infty}\mathbb{E}(f(X_n)) = \mathbb{E}(f(X)) $$
for all continuous and bounded functions $\mathbb{R}^d \to \mathbb{R}$. since $y \mapsto e^{i\langle x, y \rangle} $ is continuous and bounded for any $x\in \mathbb{R}^d$, we obtain pointwise convergence of the characteristic functions.
