Find the eigenvalues and a basis for an eigenspace of matrix A 
Find the eigenvalues and a basis for each eigenspace of matrix A:
\begin{bmatrix}
    1 & -3 & 3 \\
    2 & -2 & 2 \\
    2 & 0 & 0 \\
\end{bmatrix}

I found the eigenvalues by computing $|A-\lambda I|$:
$\lambda_1 = 0,$
$\lambda_2 = 1,$
$\lambda_3 = -2$
How do I find a basis for each eigenspace of matrix A?
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I tried the following:
$\lambda = 0:$
\begin{bmatrix}
    1 & -3 & 3 \\
    2 & -2 & 2 \\
    2 & 0 & 0 \\
\end{bmatrix}
Do reduced row echelon form:
\begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & -1 \\
    0 & 0 & 0 \\
\end{bmatrix}
Does this mean one of the results is 
\begin{bmatrix}
   0\\
   t\\
   t\\
\end{bmatrix}
and to find the other two answers, do the same thing except set $\lambda$ equal to the other two values? Or does finding the basis for each eigenspace of matrix A mean something different?
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You’ve described the general process of finding bases for the eigenspaces correctly. Note that since there are three distinct eigenvalues, each eigenspace will be one-dimensional (i.e., each eigenspace will have exactly one eigenvector in your example). If there were less than three distinct eigenvalues (e.g. $\lambda$=2,0,2 or $\lambda$=2,1), there would be at least one eigenvalue that yields more than one eigenvector. For this matrix, however, you can find the eigenvalues ($\lambda$s) and at least two corresponding eigenvectors by inspection.
Observe that the product of a matrix and column vector is a linear combination of the columns of the matrix with coefficients given by the vector. Adding the 2nd column to the 3rd produces 0 (the column vector of all 0s), so
$\begin{bmatrix}
    0\\1\\1
\end{bmatrix}$
is an eigenvector of $\lambda=0$. Because these two columns cancel each other if their coefficients in a linear combination are equal, one can also see that
$A\begin{bmatrix}
    1\\2\\2
\end{bmatrix}
=
\begin{bmatrix}
    1\\2\\2
\end{bmatrix}$, so
$\begin{bmatrix}
    1\\2\\2
\end{bmatrix}$ is an eigenvector of $\lambda=1$. You get the third eigenvalue “for free” because the trace of the matrix is equal to the sum of the eigenvalues, so the last eigenvalue is $1+(-2)-(0+1)=-2$. An eigenvector for this last eigenvalue might be harder to spot than the first two (subtract the 1st column from the 3rd), so you can always fall back on the standard method for this last eigenspace, namely, finding the null space of $A+2I$.
