Let $\Omega = \mathbb{R}$ and $\mathcal{F}$ be the collection of all finite unions of disjoint intervals of the form $(a,b]\cap\mathbb{R}$, $-\infty\leq a < b\leq \infty$. Show that $\mathcal{F}$ is an algebra but not a $\sigma$-algebra.
MY ATTEMPT
Indeed, $\Omega\in\mathcal{F}$: it suffices to take $a = -\infty$ and $b = \infty$.
If $A = (a,b]$, then $A^{c} = (-\infty,a]\cup(b,\infty)\in\mathcal{F}$ because $(-\infty,a]\in\mathcal{F}$, $(b,\infty)\in\mathcal{F}$ and the union is finite.
Finally, if $A = (a,b]$ and $B = (c,d]$, then one has that \begin{align*} A\cap B = (a,b]\cap(c,d] = \begin{cases} \varnothing & \text{if}\,\,(a\geq d)\vee(b\leq c)\\\\ (c,b] & \text{if}\,\,a \leq c < b \leq d\\\\ (a,d] & \text{if}\,\,c \leq a < d\leq b \end{cases} \end{align*} which clearly belongs to $\mathcal{F}$. But $\mathcal{F}$ is not a $\sigma$-algebra. Indeed, it suffices to consider the sequence of sets: \begin{align*} S_{n} = \left(a,b - \frac{1}{n}\right]\in\mathcal{F} \Rightarrow \bigcup_{n\in\mathbb{N}} S_{n} = (a,b)\not\in\mathcal{F} \end{align*} Consequently, $\mathcal{F}$ is not a $\sigma$-algebra.
Is there any theoretical flaw in my reasoning? Any contribution is appreciated.