Prove that the TM which would go over the leftmost position is not decidable
You can reduce the halting problem to your problem. Roughly speaking, the reduction goes along the following lines:


  Given a Turing machine $T$, construct a Turing machine $T'$ which simulates $T$ but never ventures to the left of the starting symbol. If $T$ halts, it then proceeds to go all the way to the left, until it goes beyond the starting symbol.


If you can do this, then you will have shown that LEFT is undecidable.

So how do you construct the machine $T'$? The difficult part is to somehow arrange that it never goes to the left of the starting symbol. Fortunately, this is essentially equivalent to showing that Turing machines with a one-sided infinite tape are equivalent to Turing machines with a two-sided infinite tape. You can look this up in many sources, or come up with a simulation of your own. There are many different ways to do it. Be creative!



One thing which might be unclear is what it means to simulate $T$ without going to the left of the starting symbol. What if $T$ wants to go to the left of the starting symbol? We are not looking for a literal simulation. Rather, we want to simulate each step of $T$ using a series of steps in $T'$, in such a way that the state of $T$ can be read from the state of $T'$. As a simple example, you can think of a Turing machine over the tape alphabet $\{0,1\}$ which simulates a Turing machine with tape alphabet $\{00,01,10,11\}$ by using two cells to store each symbol. We are looking for something similar: simulating a two-sided infinite tape using a one-sided infinite tape.