= Solution
For <probability measures> on a metric space, <weak convergence of probability measures> $\mu_n\Rightarrow\mu$ means $\int f\,d\mu_n\to\int f\,d\mu$ for every bounded continuous real function $f$. On the real line it is equivalently convergence of the <cumulative distribution functions> at every continuity point of the limiting <cumulative distribution function>. Bounded continuous upper and lower approximations to interval indicators give one direction; a partition at continuity points, uniform approximation on a compact interval, and control of the tails give the reverse direction.
For a <probability measure> $\mu$ on $\mathbb R$, its <characteristic function> is $\varphi_\mu(t)=\int e^{itx}\mu(dx)$. The <Lévy continuity theorem> states: $\mu_n\Rightarrow\mu$ implies $\varphi_{\mu_n}(t)\to\varphi_\mu(t)$ for every real $t$; conversely, if these <characteristic functions> have a pointwise limit $\varphi$ which is continuous at zero, there is a unique <probability measure> $\mu$ with <characteristic function> $\varphi$, and $\mu_n\Rightarrow\mu$. In particular, for a specified limiting <probability measure>, weak convergence is equivalent to pointwise convergence of its <characteristic functions>.
The forward implication follows by testing cosine and sine. The <dominated convergence theorem> shows that every <characteristic function> is continuous at zero and has value one there. For the converse, first establish <tightness of probability measures>. By <Fubini's theorem>, for $h>0$,
$$
\frac1{2h}\int_{-h}^h(1-\operatorname{Re}\varphi_{\mu_n}(t))\,dt
=\int\left(1-\frac{\sin(hx)}{hx}\right)\mu_n(dx).
$$
The integrand on the right is at least $1/2$ when $|x|\geq2/h$. This proves the <characteristic-function tightness bound>
$$
\mu_n(|x|\geq2/h)\leq\frac1h\int_{-h}^h(1-\operatorname{Re}\varphi_{\mu_n}(t))\,dt.
$$
The integrals converge for each $h$ by <dominated convergence>, and continuity of $\varphi$ at zero makes the limit as small as desired by choosing $h$ small. Thus all sufficiently late measures have a common tail bound; the finitely many earlier measures also have small tails once the radius is increased. This is uniform <tightness of probability measures>.
We give the required compactness argument explicitly. For any subsequence, choose a further diagonal subsequence whose <cumulative distribution functions> $F_n$ converge at every rational $q$, to $f(q)$. Define
$$
F(x)=\inf_{\substack{q>x\\q\in\mathbb Q}}f(q).
$$
This function is nondecreasing and right-continuous: given a rational $q>x$ with $f(q)$ close to $F(x)$, the same bound holds for $F(y)$ when $x<y<q$. Tightness gives $F(-\infty)=0$ and $F(+\infty)=1$, so $F$ is a <cumulative distribution function> of a <probability measure> $\nu$. For rationals $r<x<s$, $F_n(r)\leq F_n(x)\leq F_n(s)$. The limits bracket $F(x-)$ and $F(x)$; at a continuity point of $F$ they coincide. Thus $F_n(x)\to F(x)$ at all such points. To obtain weak convergence, choose a compact interval with small tails and partition it using continuity points of $F$. A bounded continuous function is uniformly continuous there, so step-function approximations and convergence of interval probabilities prove convergence of its integral. This proves <Helly selection for distribution functions> and supplies a weak subsequential limit $\nu$.
Its <characteristic function> is $\varphi$ by the already proved forward implication. We also prove uniqueness rather than assume it. If two <probability measures> $\nu,\rho$ have the same <characteristic function>, convolve each with a centered <normal distribution> of variance $\varepsilon^2>0$. Their densities are, by the <Gaussian integral> and <Fubini's theorem>,
$$
p_{\nu,\varepsilon}(x)=\frac1{2\pi}\int_{\mathbb R}e^{-itx}\varphi_\nu(t)e^{-\varepsilon^2t^2/2}\,dt,
$$
and likewise for $\rho$. The integrable Gaussian factor justifies exchanging the integrals, so these densities agree. For every bounded continuous $f$, letting $\varepsilon\downarrow0$ in $\int\mathbb E[f(x+\varepsilon Z)]\nu(dx)$, with $Z$ a standard <normal random variable>, gives $\int f\,d\nu$ by <bounded convergence theorem>. Hence the two measures have equal integrals against all these test functions, and approximating interval indicators shows $\nu=\rho$. This proves the <uniqueness theorem for characteristic functions>.
Every subsequence therefore has a further subsequence weakly converging to the same unique $\mu$. If the original sequence failed weak convergence, some bounded continuous test function would have a subsequence of integrals staying a fixed distance from its integral under $\mu$, contradicting that further convergence. This completes both directions of the theorem. Continuity at zero is essential: the laws $N(0,n)$ have pointwise limiting <characteristic function> zero away from zero and one at zero, and are not tight.
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