= Solution
{c}
The definition of <convergence in distribution> gives $F_n(t)\to F(t)$ at each <continuity point> $t$ of $F$. We must handle a moving argument rather than assume this convergence is uniform.
A <cumulative distribution function> has at most countably many <discontinuities>: each is a jump of positive size, and there are at most $r$ jumps of size at least $1/r$, since their sizes sum to at most one. Thus we can choose <continuity points> $u_j<c<v_j$ with $u_j\uparrow c$ and $v_j\downarrow c$. For fixed $j$, eventually $u_j<c_n<v_j$. The <monotonicity> of each <cumulative distribution function> implies
$$
F_n(u_j)\leq F_n(c_n)\leq F_n(v_j).
$$
First let $n\to\infty$, using the definition of <convergence in distribution> at $u_j,v_j$. Then let $j\to\infty$, using <continuity> at $c$:
$$
F(u_j)\leq\liminf_nF_n(c_n)\leq\limsup_nF_n(c_n)\leq F(v_j),\qquad
\boxed{F_n(c_n)\longrightarrow F(c).}
$$
No result relating different <convergence of random variables> has been assumed.
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