Deactivate all independent sources: replace the $1\,\text{V}$ source by a short and the two $1\,\text{A}$ sources by open circuits.
After this, only the resistors remain. The short across the voltage source ties the left top node to the bottom rail, so from node $a$ to $b$ there are:
\[
\text{(i) } 1\Omega + 1\Omega = 2\Omega \text{(top series path to the shorted left node)}
\]
in parallel with
\[
\text{(ii) } 1\Omega + 1\Omega = 2\Omega \text{(the rightmost vertical branch).}
\]
Hence,
\[
R_{\text{th}} = 2\Omega \parallel 2\Omega=\frac{2\times 2}{2+2}=1.0\,\Omega.
\]