Durrett's probability book appears to still be free (on author's page). Your subject is embedded in Chapter 3 Central Limit Theorems (Weak Convergence, Characteristic Functions etc., leading to Continuity Theorem 3.3.6).
Rick Durrett's Probability: Theory and Examples book Theorem 3.3.6. Levy's continuity theorem:
Let $\mu_n$, $1\leq n \leq \infty$ be probability measures with characteristic functions $\phi_n$.
(i) If $\mu_n$ converges weakly to $\mu_\infty$, then $\phi_n(t)$ converges pointwise to $\phi_\infty(t)$.
(ii) If $\phi_n(t)$ converges pointwise to a limit $\phi(t)$ that is continuous at $0$, then the associated sequence of distributions $\mu_n$ is tight and converges weakly to a measure $\mu$ with characteristic function $\phi$.
Statement (i) follows from the Portmanteau Theorem characterization of weakly convergence that uses bounded and continuous functions, by noting that $\mathrm e^{itx}$ is bounded and continuous in $x$.
For (ii), Durrett first proves that sequence of distributions $\mu_n$ is tight (not easy, uses the continuity of $\phi$ at $0$; I'll try to update some details later), which in turn implies the existence of a weakly convergent subsequence. The distribution, call it $\mu$, this subsequence converges weakly to, must have characteristic function $\phi$. Moreover, every subsequence has a further subsequence that converges weakly to $\mu$. Using again the Portmanteau Theorem characterization mentioned above (and a general topological fact: if every subsequence has a further subsequence that converges to some point, then the whole sequence converges to that point) one can show that the whole sequence of distributions $\mu_n$ converges weakly to $\mu$.
Edit: The tightness follows from the observation that:
$$ \mu_n\left(\{x\;|\;|x|>2u^{-1} \}\right) \leq u^{-1}\int_{-u}^u (1- \phi_n(t))dt,$$
using Fubini's theorem to re-write the integral as:
$$ u^{-1}\int_{-u}^u (1- \phi_n(t))dt = \int_{-\infty}^\infty \left(1- \frac{\sin(ux)}{ux}\right)\mu_n(dx),$$
and the fact that $|\sin x|\leq |x|$ for all $x$.
Edit2: For relationship of tightness and weak convergence topology see also Prokhorov's Theorem and its corollaries.
Please notice following example. Let
\begin{equation*}
F_n(x)=\frac{[n(x^+\wedge 1)]}{n}, \quad n\ge 1,\qquad F(x)=x^+\wedge 1,\quad x\in\mathbb{R}, \tag{1}
\end{equation*}
$\mu_n,\mu$ be the measures generated by $F_n,F$ on $(\mathbb{R},\mathscr{B}_{\mathbb{R}})$ respectively. Then, as $n\to\infty$,
\begin{equation*}
\mu_n\overset{w}{\to} \mu,\qquad F_n(x)\to F(x), \quad \forall x\in \mathbb{R}. \tag{2}
\end{equation*}
Also, let
\begin{equation*}
A_n=\Big\{\frac{k}{n}, k=0,1,\cdots,n\Big\},\quad n\ge1,\qquad A=\bigcup_{n\ge 1}A_n,
\end{equation*}
then $A_n,A$ are countable and
\begin{gather*}
\mu_n(A_n)=\mu_n(A)=1,\qquad n\ge 1, \tag{3}\\
\mu(A_n)=\mu(A)=0, \qquad n\ge 1. \tag{4}
\end{gather*}
(2)--(4) means that weak convergence (to a continuous distribution) does not
imply "strong convergence" of measures.
Best Answer
Hint: On any probability space $X,-X,X,-X,\cdots$ converges in distribution if and only if $X$ and $-X$ have the same distribution. For a counter-example all you need is two probability measures $P$ and $Q$ which are absolutely continuous w.r.t. each other such $X$ had a symmetric distribution w.r.t $P$ but not w.r.t $Q$. I will leave the construction to you.