Let $Y_1,Y_2,\dots$ i.i.d random variables with values in $\mathbb R^2$. Let $Z_n= \sum_{i=1}^n Y_i, n\in\mathbb N$. I want to calculate the weak limit $\frac {Z_n}{\sqrt n}$ if $Y_1\sim U\{(-1,0),(1,0),(0,1),(0,-1)\}$.
I think $\operatorname{E}(Y_1)=0$ and $\operatorname{Var}(Y_1)=1$. Then I think I have to compute the covariance matrix, it is then given by $$ \begin{pmatrix} \operatorname{Var}(Y_1) & \operatorname{E}(Y_1)\\ \operatorname{E}(Y_1) & \operatorname{Var}(Y_1) \end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} =: A$$ Therefore $\frac{Z_n}{\sqrt n}$ converges to $N(0,A)$ by central limit theorem.
Is this reasoning correct? Actually I was not sure about the calculation of $\operatorname{Var}(X_1)$, it was more a guess...
Assuming that you mean that $Y_{1}$ can only take those 4 points listed in the set $U$, the Expected Value is
$\begin{align} E[Y_{1}] = \frac{1}{4}*(-1,0) + \frac{1}{4}*(1,0) + \frac{1}{4}*(0,1) + \frac{1}{4}*(0,-1) = 0 \end{align}$
and because the expected value is $0$ the variance is the 2nd moment.
But let's be a little more precise with our definition of $Y_1$ so we can talk about the variance-covariance matrix. We have that $Y_1 = (y_1, y_2)$ where you can think of $y_1,y_2$ as bernoulli(1,-1) random variables. If we take the expected value of a random vector we get
$E[Y] = (E[y_1], E[y_2]) = (0,0)$
and $y_1$ takes values 1,-1,0,0 with probability $\frac{1}{2}$ each. The variance matrix of the random vector $X = (X_1,...,X_n)$ is the $nxn$ matrix defined as
$[Var(X)]_{ij} = Cov(X_i, X_j)$
If we calculate $Cov(y_1, y_1)=Var(y_1) = \frac{1}{4}(-1)^2 + \frac{1}{4}(1)^2 + \frac{1}{2}(0) + \frac{1}{2}(0) = \frac{1}{2}$
And the same thing works for $Cov(y_2,y_2)$
Then if we calculate $Cov(y_1, y_2) = E[y_1y_2]-E[y_1]E[y_2] = E[y_1y_2]$ which is clearly $0$ because the coordinates always have one being $0$ and the other being $1$ or $-1$.
Then its easier to write the covariance matrix $A$ as \begin{pmatrix} \operatorname{Var}(y_1) & \operatorname{Cov}(y_1, y_2)\\ \operatorname{Cov}(y_2, y_1) & \operatorname{Var}(y_2) \end{pmatrix} =\begin{pmatrix} \frac{1}{2}& 0\\ 0 & \frac{1}{2} \end{pmatrix}
Which is the matrix you need for $A$
So now use the fact that the $Y_i$'s are independent and take $Var(Z_n) = Var(\sum_{i=1}^n Y_i) = \sum_{i=1}^n Var(Y_i) = nA$
Then looking at the quantity $\frac {Z_n - 0}{\sqrt n}$ we can easily see that $E[\frac {Z_n}{\sqrt n}] = 0$ and that $Var(\frac {Z_n - 0}{\sqrt n}) = \frac{1}{n}Var(Z_n) = \frac{nA}{n} = A$ so then applying the central limit theorem gives you $\frac {Z_n}{\sqrt n}$ goes to a two dimensional normal $N_2(0,A)$ as $n$ goes to infinity. So yes your reasoning is correct!