Proving there are infinitely many strictly dominant strategies My attempt: 

$(0, 1/3, 2/3)$ works as a strategy that strictly dominated $L$. 

To prove that there are infinitely many of them:

Let $A_n=(0,1/2+\epsilon_n,1/2-\epsilon_n)$ where $\epsilon_n=1/n$ for $n=1,2,3,...$. We can see that for $n &gt; 10$, $A_n$ strictly dominates L. So, since $n$ diverges, there are infinitely many $n$ for which $A_n$ is a strictly dominant strategy.

I chose $n&gt; 10$ since $2&lt;2.5-5\epsilon_n \Leftrightarrow 0,1&gt; \epsilon_n$ and $1/n&lt;0,1$ for $n&gt;10$. The other inequalities don't require $\epsilon_n$ to be restricted. 

Would this be a correct way to prove this?
That is a fine proof. Though you probably want to show your last statement "The other inequalities don't require $\epsilon_n$ to be restricted". It is fairly obvious, but won't hurt to be super clear.

Perhaps a simpler proof would be that any convex combination of the two strategies already found will also work. That is $\lambda(0,1/2,1/2)+(1-\lambda)(0,1/3,2/3)$ also strictly dominates $L$ for any $\lambda\in(0,1)$. So, in fact, there are not only countably many strategies that strictly dominate $L$ but uncountably many of them. 

(Idea of the proof:) Notice that this holds because strict dominance is given by a set of inequalities (3 in this example) so if you have two strategies that satisfy those inequalities, any convex combination of these strategies will necessarily also satisfy the inequalities since expected utility is linear in probabilities. I'll leave the formal proof for you to write, but hopefully the logic is clear.