To show that these strategies form a (Nash) equilibrium you must show that each agent would not like to deviate from their strategy taking as given the other agent's strategy.
Agent $1$
If agent $2$ is playing
$$b_2(v_2)=\begin{cases}
v_2&\text{ if } &v_2\leq 1\\
2&\text{ if }&v_2>1
\end{cases}$$
agent $1$ has to consider three cases:
$[1]$ If $v_2\le 1$ and $v_1 \lt v_2$, then $b_2=v_2$ and following the strategy agent $1$ loses the auction getting a payoff of $0$. The only relevant deviation is to bid $b_1>v_2$ which would lead to a payoff of $v_1-v_2\lt 0$.
$[2]$ If $v_2\le 1$ and $v_1 = v_2$, then $b_2=v_2$ and following the strategy there is a tie. What happens in this case is not specified, but in any case it does not matter since if agent $1$ gets the item or not the payoff will be $v_1-v_2=0$.
$[3]$ If $v_2\le 1$ and $v_1 \gt v_2$, then $b_2=v_2$ and following the strategy agent $1$ wins the auction getting a payoff of $v_1-v_2>0$. Bidding any other $b_1>v_2$ would lead to exactly the same payoff, hence does not justify a deviation. Finally, bidding $b_1\le v_2$ would lead the agent to lose the auction getting a payoff $0$.
$[*]$ Notice that $v_2>1$ is ruled out by assumption.
Agent $2$
If agent $1$ is playing
$$b_1(v_1)=\begin{cases}
v_1&\text{ if } &v_1\leq 1\\
1&\text{ if }&v_1>1
\end{cases}$$
agent $2$ has to consider the same three cases:
$[1]$ If $v_2\le 1$ and $v_1 \lt v_2$, then $b_1=v_1$ and following the strategy agent $2$ wins the auction getting a payoff of $v_2-v_1>0$. Bidding any other $b_2>v_1$ would lead to exactly the same payoff, hence does not justify a deviation. Finally, bidding $b_2\le v_1$ would lead the agent to lose the auction getting a payoff $0$.
$[2]$ If $v_2\le 1$ and $v_1 = v_2$, then $b_1=v_1$ and following the strategy there is a tie. Again, what happens in this case is not specified, but in any case if agent $2$ gets the item or not the payoff will be $v_1-v_2=0$.
$[3]$ If $v_2\le 1$ and $v_1 \gt v_2$, then $b_1>b_2$ and following the strategy agent $2$ loses the auction getting a payoff of $0$. The only relevant deviation is to bid $b_2>\min\{v_1,1\}$ which would lead to a payoff of $v_2-\min\{v_1,1\}\le 0$.
Efficiency
Notice that the agent that values the item the most always wins the auction so the resulting allocation is efficient.
