The following proof agrees with the lines in Eliran H's answer, but it provides the justification.
The OP has the following concern which I would like to address by explaining how not to get stuck in the subproofs:
I want to start with an assumption of (A&B) and try to isolate a subproof where I can then get B⊃C through a conditional introduction, but I keep getting stuck in my subproofs and unable to reason out of them.
Here is the result using Klement's proof checker. Note that I am using "∧" rather than "&" and "→" rather than "⊃" so that I have well-formed formulas recognizable by the proof checker I am using. Also the inference rules are those the proof checker accepts. You may be required to use some different terminology:
On line 1 is the premise: "(A ∧ B) → C"
Because I will have two conditionals, rather than assuming "(A&B)", I will break this up into two subproofs, one assuming "A" and the other assuming "B". This will allow me to get two conditionals.
I start the first subproof on line 2. Note how the subproof is indented in this style of presentation. I immediately start the second subproof on line 3. At this point I have both "A" (line 2) and "B" (line 3) and so I can combine them with conjunction introduction (∧I) on line 4.
This is the first line of the proof that was not a premise nor an assumption. I need to provide a justification for it. That justification is given by "∧I 2, 3". Those symbols mean I am using conjunction introduction on lines 2 and 3.
Since line 4 is the premise needed to get "C" on line 1, I can use modus ponens or conditional elimination (→E) as the proof checker I am using requires me to call it and reference lines 1 and 4 for the justification.
At this point I can close the subproof starting on line 3. This discharges the assumption on line 3. I assumed "B" and reached a line that has "C" on it so I can write on line 6 "B→C" which, if you think about it is just another way of writing that subproof--if I assume "B" then I can derive "C".
Note that I can now do the same thing for the subproof starting at line 2 and ending at line 6 again using conditional introduction. When I do that I get the desired result. The proof checker notes this with an acknowledgement that the proof is correct.
For more information about these particular rules see forall x. The link to the proof checker provides directions for how you may use it to check your other proofs.
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/