If $X$ and $Y$ are independent and have continuous cdfs, then $X-Y$ has a continuous cdf. My question is this:

Prove that if $X$ and $Y$ are independent random variables with continuous cdfs, then $X-Y$ also has a continuous cdf.

I can prove this in many ways if I am allowed to use the fact that $X$ and $Y$ have densities. Then I can say that $X$ and $-Y$ are independent and I can use convolution to arrive at a density of $X-Y$. Or I can use the Jacobian formula and arrive at a density .
But I am to use the fact that a random variable is continuous if it's cdf is continuous and that's it. If it has density then it is absolutely continuous. So I am not allowed to use the density argument.
So if I am to show that $X-Y$ is continuous then I am to basically prove that $F_{X-Y}(a)-F_{X-Y}(a^{-})=0\,\,\,\forall a\in\mathbb{R}$
This is what I have tried.
Let $h>0$ be an arbitrary positive real number .
$$F_{X-Y}(a^{-})=\lim_{h\to 0}F_{X-Y}(a-h)=\lim_{h\to 0}P(X-Y\leq a-h)$$
How do I proceed?. Any help is appreciated. I am unfamiliar most of measure theoretic approach to probability, so please try and explain in basic analysis terms.
 A: Suppose $X-Y$ does not have a continuous c.d.f. Then for some value $c$ in the support of $X-Y,$ we have $\Pr(X-Y=c)>0.$ Then from the law of total probability we get
$$
0<\Pr(X-Y=c) = \operatorname E(\Pr(X-Y=c\mid Y)) = \operatorname E(\Pr(X= c+Y\mid Y))
$$
The conditional probability $\Pr(X=c+Y\mid Y)$ is a random variable that is completely determined by $Y.$ Since its expected value is positive, then there is some value $y$ of $Y$ for which it is positive. So $\Pr(X=c+y)>0.$ Thus the c.d.f. of $X$ has a discontinuity at $c+y.$
A: It suffices to assume that the cdf of $X$, $F_X$ is continuous. Take a sequence $\{a_n\}$ converging to $a$ from the left. Then,
\begin{align}
\lim_{n\to\infty}\mathsf{P}(X-Y\le a_n)&=\lim_{n\to\infty} \mathsf{E}[\mathsf{P}(X\le a_n+Y\mid Y)] \\
&= \mathsf{E}\!\left[\lim_{n\to\infty}\mathsf{P}(X\le a_n+Y\mid Y)\right]
\end{align}
by the Dominated Convergence Theorem.
Now,
$$
\mathsf{P}(X\le z+Y\mid Y)=F_X(z+Y) \quad\text{a.s.}
$$
Since $F_X$ is continuous,
$$
\lim_{n\to\infty}F_X(a_n+Y)=F_X(a+Y) \quad\text{a.s.},
$$
and thus,
$$
\mathsf{E}\!\left[\lim_{n\to\infty}\mathsf{P}(X\le a_n+Y\mid Y)\right]=\mathsf{P}(X-Y\le a).
$$
