This is basically a problem of recursive counting. Start with the uncoupled basis state, i.e. the set of states of the form
$\vert \ell m_\ell \rangle \vert s m_s\rangle$. There are clearly $(2\ell+1)(2s+1)$ of these, and the job is to reorganize them.
The key counting result is based on the observation that
$\vert \ell m_\ell\rangle\vert sm_s\rangle$ is an eigenstate of $\hat J_z=\hat L_z+\hat S_z$ with eigenvalue $M=m_\ell+m_s$. With this in mind organize your $\vert \ell m_\ell\rangle\vert sm_s\rangle$ states so that those with the same value of $M$ are on the same line. Explicitly, for instance, you would have
\begin{align}
\begin{array}{rlll}
M=\ell+s:&\vert \ell \ell\rangle\vert s s\rangle \\
M=\ell+s-1:& \vert \ell,\ell-1\rangle\vert ss\rangle&
\vert\ell\ell\rangle \vert s,s-1\rangle\\
M=\ell+s-2:&\vert \ell,\ell-2\rangle \vert ss\rangle & \vert\ell,\ell-1\rangle\vert s,s-1\rangle&\vert \ell \ell\rangle \vert s,s-2\rangle\\
\vdots\qquad& \qquad\vdots
\end{array}
\end{align}
and replace each state with a $\bullet$ to get
$$
\begin{array}{rlll}
\ell+s:&\bullet \\
\ell+s-1:&\bullet & \bullet\\
\ell+s-2:&\bullet&\bullet&\bullet\\
\vdots\qquad&\vdots
\end{array}
$$
Now, if $M=\ell+s$ is the largest value and it occurs once, the value of $j=\ell+s$ must occur once and also all the states $\vert j=\ell+s,m_j\rangle$ will occur once. There is a linear combination of the two states with $M=\ell+s-1$ that will be the state $\vert j=\ell+s,m_j=\ell+s-1\rangle$, there will be a linear combination of the three states with $M=\ell+s-2$ that will be the $\vert j=\ell+s,m_j=\ell+s-2\rangle$ state etc. Since we are only interested in enumerating the possible resulting values of $j$, and not interested in the actual states per se, we can eliminate from our table the first column since it contains one state with $m_j=\ell+s$, one with $m_j=\ell+s-1$ etc. Eliminating this column yields the reduced table
$$
\begin{array}{rll}
\ell+s-1: & \bullet\\
\ell+s-2:&\bullet&\bullet\\
\vdots\qquad&\vdots
\end{array}
$$
Since the value of $m_j=\ell+s-1$ occurs once, the value $j=\ell+s-1$ must occur once, and the states $\vert \ell+s-1,m_j\rangle$ will each occur once. We take out those from the list by deleting the first column to obtain a further reduced table
$$
\begin{array}{rl}
\ell+s-2:&\bullet\\
\vdots\qquad &\vdots
\end{array}
$$
The process so continues until exhaustion. In the examples above we have found $j=\ell+s,\ell+s-1$ and the final reduced table of the example, if not empty, would indicate the value of $j=\ell+s-1$. It is clear this process produces a decreasing sequence of $j$. The last value of $j$ is determined by the width of the original table. It is not hard to convince yourself that the width of the table will stop increasing once we reach $M=\vert \ell-s\vert $, and this is the last value of $j$. Thus by exhaustion you find the possible values of $j$ in the range
$$
\vert \ell-s\vert\le j\le \ell+s\, .
$$
As an example consider $\ell=1$ and $s=2$. The original table then looks like
$$
\begin{array}{rlll}
\frac{3}{2}:&\vert 11\rangle\vert 1/2,1/2\rangle \\
\frac{1}{2}:&\vert 10\rangle\vert 1/2,1/2\rangle & \vert 11\rangle\vert 1/2,-1/2\rangle\\
-\frac{1}{2}:&\vert 1,-1\rangle\vert 1/2,1/2\rangle&\vert 10\rangle\vert 1/2,-1/2\rangle\\
-\frac{3}{2}:&\vert 1,-1\rangle \vert 1/2,-1/2\rangle
\end{array}\qquad \to \qquad
\begin{array}{rlll}
\frac{3}{2}:&\bullet \\
\frac{1}{2}:&\bullet & \bullet \\
-\frac{1}{2}:&\bullet &\bullet \\
-\frac{3}{2}:&\bullet
\end{array}
$$
It is only $2$ column wide, and the width stop growing at $M=1/2$, indicating the possible $j$ in this case are $3/2$ and $1/2$, and indeed
$$
\vert 1-1/2\vert \le j\le 1+1/2
$$
Finally, note that the absolute value is required on the left because one could write state $\vert sm_s\rangle\vert \ell m_\ell\rangle$ without affecting the possible values of $j$.
