The dihedral group of order $2n$ is generated by rotation $x$ through an angle $\frac{2\pi}{n}$ and a "flip" or reflection $y$. (These satisfy the relations $x^n=y^2=1$ and $yx^i=x^{-i}y$). The subgroup generated by $x$ is isomorphic to $(\mathbb{Z}/n\mathbb{Z},+)$, which has a cyclic subgroup or order $d$ for each positive divisor $d$ of $n$. Likewise, corresponding to each of these cyclic subgroups is a dihedral subgroup of order $2d$, the symmetry group of the $d$-gon. These subgroups can also be represented by the generators:
$$
\left<x^{n/d}\right> \cong C_d \cong \left(\mathbb{Z}/d\mathbb{Z},+\right)
$$
$$
\left<x^{n/d},~y\right> \cong D_d
$$
Thus, up to isomorphism, there are $2\tau(n)$ subgroups of $D_n$. However, the union of each coset $y\left<x^{n/d}\right>x^i=\{yx^{i+jk}|0\leq k<d\}$ with $\left<x^{n/d}\right>$ , for $0\leq i<k=n/d$, is isomorphic to $D_d$, forming a distinct dihedral subgroup of $D_n$ within each isomorphism class (this is Yaghoub Sharifi's Lemma 3 in @sergey-filkin's link). Since for each $d|n$ there are $n/d$ such cosets, we get $\sum_{d|n}\frac{n}d=\sum_{d|n}d=\sigma(n)$ different dihedral subgroups and $\tau(n)$ different cyclic subgroups of $D_n$, or $\tau(n)+\sigma(n)$ total subgroups of $D_n$.
Your thinking is more or less correct. Generally, when we wish to think about the subgroups of a group $G$, we take different subsets of $G$ and look at what groups those subsets generate.
The obviously thing to do is start with one element. Let's say I start with a rotation $r^k$ for some $k = 1, 2, 3, 4$. Since $5$ is a prime, the numbers $1, 2, 3, 4$ have multiplicative inverses mod $5$. This implies that $\langle r^k\rangle$ actually contains $r$. Hence, $\langle r^k\rangle = \langle r\rangle = \{1, r, r^2, r^3, r^4\}$ for each $k$. Now, let's say I start with a reflection $z$ instead. It's easy to see that $\langle z\rangle = \{1, z\}$. Thus, we have $5$ different subgroups of order $2$, one for each reflection. The only remaining case is $\langle 1\rangle = \{1\}$.
Now let's consider a generating set with two elements. If we take $\langle r^k, r^l\rangle$ for some $k, l$, this is clearly just $\langle r\rangle$ again. Suppose instead we take $\langle r^k, z\rangle$. Then this subgroup contains $r$ and $z$, hence it contains $r^l z$ for each $l=0, 1, 2, 3, 4$, i.e. it contains every reflection. It also contains the powers of $r$, which are the rotations, hence $\langle r^k, z\rangle = G$. Finally, we could take $\langle z_1, z_2\rangle$ for two reflections $z_1, z_2$. But then $z_1 z_2$ is a rotation which is in $\langle z_1, z_2\rangle$, hence by the previous case we have $\langle z_1, z_2\rangle = G$.
From here it's easy to see that we've found every subgroup of $G$. Obviously this took a lot less time than checking all $2^{10}$ subsets of $G$.
Best Answer
Normally Hasse diagrams of lattices are drawn so that "big things" are at the top, and a line between items indicates that there is no other node between those two items.
Here's a start (you'll have to complete it)
The dihedral group for the pentagon is not really very fun since there are so few divisors of $10$. You will get a much more interesting exercise if you try the dihedral group for the hexagon. I encourage you to try it out!