For working out the Thevenin resistance, your approach was wrong.
Look at the following schematics which I believe accurately describe your approach to find the Thevenin resistance. With \$R_{_\text{L}}\$ disconnected and \$V_1\$ shorted, you transition from the left-side to the right-side, below:

simulate this circuit – Schematic created using CircuitLab
Here, you take \$R_1\$ in parallel with \$R_2\$, first. That combination will then be in series with \$R_3\$. And that combination will be in parallel with \$R_4\$.
So the Thevenin resistance, seen by \$R_{_\text{L}}\$, will be:
$$R_{_\text{TH}}=\left[\left(R_1\mid\mid R_2\right)+R_3\right]\mid\mid R_4$$
(You should be able to see that \$R_4\$ is directly in parallel with \$R_{_\text{L}}\$, anyway. So you should know, right away, that whatever answer you come up with -- that this answer must end with putting \$R_4\$ in parallel with the immediately prior result.)
For working out the Thevenin voltage, your approach that got \$5.44\:\Omega\$ is workable.

simulate this circuit
First, sum up \$R_3\$ and \$R_4\$, as you indicated. Then put that result in parallel with \$R_2\$. That will, in fact, result in \$R_{_\text{A}}=\left(R_3+R_4\right)\,\mid\mid\, R_2=5.44\:\Omega\$.
\$R_{_\text{A}}\$ now makes a voltage divider with \$R_1\$, with the resulting voltage being \$V_{_\text{A}}=V_1\cdot\frac{R_{_\text{A}}}{R_1+R_{_\text{A}}}\$.
You now just want the voltage across \$R_4\$, which will be less than \$V_{_\text{A}}\$. But \$R_3\$ and \$R_4\$ are just one more voltage divider, though you want the upper-half voltage (across \$R_4\$), so this will be \$V_{_\text{TH}}=V_{_\text{A}}\cdot\frac{R_4}{R_3+R_4}\$.
So the Thevenin voltage, seen by \$R_{_\text{L}}\$, will be:
$$\begin{align*}
V_{_\text{TH}}&=V_1\cdot\frac{R_{_\text{A}}}{R_1+R_{_\text{A}}}\cdot\frac{R_4}{R_3+R_4}
\\\\
&= V_1\cdot\frac{\left(R_3+R_4\right)\,\mid\mid\, R_2}{R_1+\left[\left(R_3+R_4\right)\,\mid\mid\, R_2\right]}\cdot\frac{R_4}{R_3+R_4}
\end{align*}$$