If $X_n\overset{d}{\to}c$ for some constant $c$, does that implies that $X_n\overset{\text{a.s.}}{\to}c$? I know that if $X_n\overset{d}{\to}c$ for some constant $c$, that implies that $X_n\overset{p}{\to}c$, and that means that for every $\varepsilon>0$, $$P(|X_n-c|>\varepsilon)\to0$$ when $n\to 0$, but is there a way to deduce that $X_n\overset{\text{a.s.}}{\to}c$? Or is that simply false?

It's false.

Suppose we have a sequence of random variables $X_n$ along with $Y$ such that

$X_n - Y \to 0$ in probability. Then if your statement was correct, this would imply that

$X_n - Y \to 0$ almost surely.

But this implies

$P(\lim |X_n - Y - 0| > \epsilon )=0$

Which implies

$P(\lim |X_n - Y|> \epsilon )=0$

So that convergence in probability would imply almost sure convergence, which is obviously false.