We ought to be a little concerned about using "$\epsilon$" for two separate purposes. So, let's back up a little and generalize your idea.
Fix numbers $x,y.$ Looking at your original quotient,
suppose for all $\delta \gt 0,$ $\Pr(Y \in (y, y+\delta])$ is nonzero.
This permits us to use an elementary definition of conditional probability as
$${\Pr}_{X\mid Y}(X \in \mathcal{A} \mid Y \in (y, y+\delta]) = \frac{{\Pr}_{X,Y}(X\in \mathcal{A}, Y \in (y, y+\delta])}{{\Pr}_{X,Y}(Y \in (y, y+\delta])}\tag{*}$$
where $\mathcal A \times \mathbb{R}$ is any measurable set.
By definition, the CDF of $(X,Y)$ is
$$F(a,b) = F_{X,Y}(a,b) = \Pr(X \le a, Y \le b)$$
for any real numbers $a,b.$ Apply this in the case $\mathcal{A} = (x, x+\epsilon]$ to re-express the fraction in $(*)$ as
$$\eqalign{
{\Pr}_{X\mid Y}(\mathcal{A} \mid (y, y+\delta]) &= \frac{{\Pr}_{X,Y}(X\in (x,x+\epsilon], Y \in (y, y+\delta])}{{\Pr}_{X,Y}(Y \in (y, y+\delta])} \\
&= \frac{F(x+\epsilon,y+\delta) - F(x,y+\delta) - (F(x+\epsilon, y) - F(x,y))}{F_Y(y+\delta) - F_Y(y)}.
}$$
This is perfectly fine because we have assumed the denominator is nonzero. But we want to take the limit as $\delta$ shrinks to zero and this might be undefined. To proceed,
assume $F$ is continuously differentiable in its second argument with derivative $D_2F.$
This permits us to apply L'Hopital's Rule to the $\delta$ limit, allowing us to set
$$\eqalign{{\Pr}_{X\mid Y}(\mathcal{A} \mid Y=y) & := \lim_{\delta\to 0^+} {\Pr}_{X\mid Y}((x,x+\epsilon] \mid (y, y+\delta]) \\
&=\lim_{\delta\to 0^+} \frac{F(x+\epsilon,y+\delta) - F(x,y+\delta) - (F(x+\epsilon, y) - F(x,y))}{F_Y(y+\delta) - F_Y(y)} \\
&= \lim_{\delta\to 0^+} \frac{D_2F(x+\epsilon,y) - D_2F(x,y) - (0 - 0)}{D_2F_Y(y) - 0} \\
&= \frac{D_2F(x+\epsilon,y) - D_2F(x,y)}{D_2F_Y(y)}.
}$$
This may look strange, but let's proceed further and
suppose the function $x \to D_2F(x,y)$ is continuously differentiable at $x.$
This permits us to differentiate the preceding with respect to $x$ by taking the limit of the difference quotient thus:
$$\eqalign {
f_{X\mid Y}(x,y) & := \lim_{\epsilon\to 0^+} \frac{{\Pr}_{X\mid Y}((-\infty,x+\epsilon] \mid Y=y) - {\Pr}_{X\mid Y}((-\infty,x] \mid Y=y)}{\epsilon} \\
&=\lim_{\epsilon\to 0^+} \frac{{\Pr}_{X\mid Y}((x,x+\epsilon] \mid Y=y)}{\epsilon} \\
&= \lim_{\epsilon\to 0^+}\frac{1}{\epsilon} \left(\frac{D_2F(x+\epsilon,y) - D_2F(x,y)}{D_2F_Y(y)}\right) \\
&= \lim_{\epsilon\to 0^+}\frac{1}{D_2F_Y(y)} \left(\frac{D_2F(x+\epsilon,y) - D_2F(x,y)}{\epsilon}\right) \\
&= \frac{D_1 D_2 F(x,y)}{D_2F(x,y)} \\
&= \frac{f_{X,Y}(x,y)}{f_Y(y)}
}$$
in terms of the (conventional) shorthand $f_{X\mid Y}(x,y)$ for the conditional density of $X \mid Y=y,$ $f_{X,Y} = D_1 D_2 F_{X,Y},$ and $f_Y(y)=\mathrm{d}F_Y(y)/\mathrm{d}y:$ QED.
(The only reason to take limits through positive values of $\delta$ and $\epsilon$ and to assume continuous differentiability was due to the interval notation; the limits can be evaluated through negative values of $\delta$ and $\epsilon$ using the same method and comparable assumptions.)