Since you know that convergence in probability and convergence in distribution is equivalent if the limit is a constant, then the best way is to go to subsequential arguments.
$A_{n}\xrightarrow{d}a\iff A_{n}\xrightarrow{P}a$ .
Now assume $a\neq 0$ as otherwise the question does not make any sense really.
Let $\dfrac{1}{A_{n_{k}}}$ be an arbitrary subsequence. Then as $A_{n_{k}}\xrightarrow{P} a$, there exists a subsequence $A_{n_{k_{l}}}\xrightarrow{a.s.} a$.
Hence now by just usual properties of sequences, $\dfrac{1}{A_{n_{k_{l}}}}\xrightarrow{a.s.}\dfrac{1}{a}\implies\dfrac{1}{A_{n_{k_{l}}}}\xrightarrow{d}\dfrac{1}{a}$ .
Now here's an excercise that you should do by yourself:-
$X_{n}$ converges in distribution to a random variable $X$ if and only if given any subsequence $X_{n_{k}}$ there exists a further subsequence $X_{n_{k_{l}}}$ that converges in distribution to $X$. You will need to use this result to check that $F_{n}(x)\to F(x)$ for all continuity points $x$ of $F$.
But what the hell, here is the proof. It is clear that if
$X_{n}\xrightarrow{d}X$, then $F_{n}(x)\to F(x)$ for all continuity
points of $F$ and hence for each subsequence $F_{n_{k}}(x)$, all it's
subsequences $F_{n_{k_{l}}}(x)$ will converge to $F(x)$ for each
continuity point $x$. Conversely, by the condition, given any
subsequence $F_{n_{k}}(x)$, there exists a further subsequence
$F_{n_{k_{l}}}(x)$ which converges to $F(x)$ for each continuity point
$x$. Hence by the real analysis result I linked above, the whole
sequence $F_{n}(x)\to F(x)$ for each continuity point $x$ of $F$. Thus
$X_{n}\xrightarrow{d} X$
.
Couple the argument above and the above excercise to complete your proof.
