What is the correct metric to use for two dimensional de Sitter? If one starts with the following metric, which looks similar to de Sitter in 4 dimensions:
$$ds^2 = -dt^2 + e^{2H t} dx^2,$$
one can calculate $R = 2H^2$, and $R_{00} = -H^2$, which gives the $\Lambda = 0$, which is not the solution one is looking for. What should be the correct metric to use for the same?
In two-dimensional spacetime, the Einstein tensor $R_{ab}-\frac{1}{2}g_{ab}R$ is identically zero , which explains why you get $\Lambda=0$.
In any number $D$ of spacetime dimensions, including $D=2$, de Sitter spacetime can be constructed like this. Start with the $D+1$ dimensional Minkowski metric $$ -(\mathrm dX^0)^2+\sum_{k=1}^D(\mathrm dX^k)^2. \tag{1} $$ The submanifold defined by the condition $$ \sum_{k=1}^D(X^k)^2=L^2+(X^0)^2 \tag{2} $$ is $D$-dimensional de Sitter spacetime. The length parameter $L$ is related to the cosmological constant $\Lambda$ by $$ \Lambda = \frac{(D-2)(D-1)}{2L^2}. \tag{3} $$ This is equation (4) in "Les Houches Lectures on de Sitter Space". Setting $D=2$ recovers your result $\Lambda=0$.
By the way, equations (13)-(14) in the same paper show how to derive the de Sitter metric in the form $$ -\mathrm dt^2+e^{2t}\sum_{k=1}^{D-1}(\mathrm dx^k)^2 $$ starting from equations (1)-(2). For $D=2$, this reduces to the form shown in the question.
