Give an example of two discrete random variables X and Y on the same sample space such that X and Y have the same distribution, 
Give an example of two discrete random variables X and Y on the same sample space
  such that X and Y have the same distribution, with support {1, 2, . . . , 10}, but the event X = Y never occurs. If X and Y are independent, is it still possible to construct such an example?

My answer: It's not possible. 
Explanation: By definition of independence says that knowing about the occurrence of one of the events doesn't affect the probability of occurrence of the other. In this case $X=k$, $k \epsilon${${1, 2, . . . , 10}$} effectively reduces the sample space for $Y$ because $X=k$ means Y can take only one of the other 9 values. Hence $P(Y=y)$ is not unaffected by the occurrence of $X=k$. Hence X & Y cannot be independent. 
Is this correct?
 A: Try sample space $\{1,\ldots,10\}$ with equal probabilities, $X(i) = i$ and $Y=X+1$ when $X<10$, $1$ when $X=10$.
If $X$ and $Y$ are independent, $X=Y=i$ has positive probability whenever $X=i$ and $Y=i$ do.
A: 
A. Give an example of two discrete random variables $X$ and $Y$
  on the same sample space such that $X$ and $Y$ have the same distribution,
  with support {1, 2, . . . , 10}, but the event $X = Y$ never occurs.

An example similar to Robert Israel’s:
Choose any symmetric probability distribution for $X$;
i.e., $P(X=1)=P(X=10)$,  $P(X=2)=P(X=9)$, etc. 
A trivial example is the uniform distribution:
$P(X=i)=0.1~\forall i\in \{1, 2, . . . , 10\}$. 
A slightly more interesting example is the triangular distribution:
\begin{align}
P(1)&=.033333333&P(6)&=.166666667\\
P(2)&=.066666667&P(7)&=.133333333\\
P(3)&=.1&P(8)&=.1\\
P(4)&=.133333333&P(9)&=.066666667\\
P(5)&=.166666667&P(10)&=.033333333
\end{align}
    
Then set $Y=11-X$.
Argument: $X$ and $Y$ are distributed identically,
because, $\forall i,~P(Y=i) = P((11-X)=i) = P(X=(11-i)) = P(X=i)$
(from the symmetry of $X$’s distribution).
$X=Y \implies X=11-X \implies 2X=11 \implies X=5.5$, 
but $P(X=5.5)~=~0$,  so this never happens.

B. If $X$ and $Y$ are independent,
  is it still possible to construct such an example?

No.
Argument:
There must be some $i$ ($i\in \{1, 2, . . . , 10\}$) for which $P(X=i)>0$. 
Now
\begin{align}
P(X=i,\,Y=i)&=P(X=i)\times P(Y=i)&\text{by independence}\\
&=P(X=i)\times P(X=i)&\text{by identical distribution}\\
&=P(X=i)^2\\
&>0&\text{because $P(X=i)>0$}
\end{align}
so the event $X=Y$ occurs with positive probability.
