Step 1: Set up the cell.
In water, $NaCl$ splits into $Na^+$ and $Cl^-$, and water gives a few $H^+$ and $OH^-$ ions. So the solution has $Na^+, H^+$ moving to the cathode and $Cl^-, OH^-$ moving to the anode.
Step 2: Look at the cathode (reduction).
Two ions could be reduced here, $Na^+$ and $H^+$. Sodium is very hard to deposit from water because $H^+$ is reduced much more easily. So hydrogen gas is released, not sodium metal.
\[ 2H^+ + 2e^- \rightarrow H_2 \]
Step 3: Look at the anode (oxidation).
Here $Cl^-$ and $OH^-$ compete. Because the chloride is in fairly high concentration (brine), chlorine gas comes off in practice.
\[ 2Cl^- \rightarrow Cl_2 + 2e^- \]
Step 4: What is left in solution.
Once $H^+$ is pulled out as $H_2$, the leftover $OH^-$ stays with $Na^+$ in the beaker. That means $NaOH$ builds up in the solution.
Step 5: Collect all products.
At the electrodes we get $H_2$ and $Cl_2$ gases, and the solution becomes $NaOH$.
Step 6: Pick the answer.
So the products are $NaOH, H_2$ and $Cl_2$ together.
\[ \boxed{NaOH,\ H_2,\ Cl_2} \]