Is this proof correct? (natural deduction)

You were right to doubt your proof; it's not quite right.

The main mistake is that you are effectively closing two subproofs at once once you go from 2.2.3 to 3, but you can only close one subproof at a time, in the reverse order in which you opened them.

Also: you need to have line 1 as an assumption of a subproof, and close that subproof once you have reached line 3, so you can infer the whole desired conditional by $\rightarrow$ Intro

So, what to do? Well, that depends on the rules that you have. Let me give you some options:

Option 1:

After 2.2.3 you can get:

2.3 $\neg Q \rightarrow Q$

Then use the Implication equivalence to get $\neg \neg Q \lor Q$ and thus by Double Negation to $Q \lor Q$, and finally by Idempotence to $Q$

Option 2:

First, get $Q \lor \neg Q$ by Law of Excluded Middle. Then again get $\neg Q \rightarrow Q$, but do a second inside subproof where you assume $Q$, reiterate $Q$, close that subproof to get $Q \rightarrow Q$, and then get $Q$ from $\neg Q \rightarrow Q$, $Q \rightarrow Q$, and $Q \lor \neg Q$ by a proof by cases, probably formalized as $\lor $ Elim

Option 3:

Get a contradiction on 2.2.4 between $\neg Q $ and $Q$, and thus conclude $\neg \neg Q$ by a proof by contradiction (probably formalized by $\neg \ Intro$), and thus the $Q$ you really want.

I would do it slightly differently, but as always people use somewhat different systems for natural deduction.

• Begin
  • 1) $(P \land \lnot Q) \rightarrow Q$ (ass.)
    • 2) $P$ (ass.)
      • 3) $\lnot Q$ (ass.)
      • 4) $(P \land \lnot Q)$ introduction $\lnot$ from 2,3.
      • 5) $Q$ (modus ponens from 1. and 4.)
      • 6) $\bot$ from $5$ and $3$.
    • 7) $\lnot \lnot Q$ (from introduction $\lnot$ and 3, 6, we drop 3).
    • 8) $Q$ (double negation rule).
  • 9) $P \rightarrow Q$ (from 2,8 and introduction $\to$); we drop 2.
• 10) $((P \land \lnot Q) \to Q)\to (P \to Q)$ (intro $\to$ 1, 9) we drop 1.

Done.

$\def\fitch#1#2{\begin{array}{|l}#1\\\hline#2\end{array}}$

Can you use the Law of Excluded Middle?

$$\fitch{}{\fitch{(P\land\lnot Q)\to Q}{\fitch{P}{\left.\raise{5ex}{Q\vee\lnot Q\\\fitch{Q}{Q}\\Q\to Q}\right\}&\text{add this}\\\fitch{\lnot Q}{P\wedge\lnot Q\\ Q}\\\lnot Q \to Q\\Q}\\P\to Q}\\((P\land\lnot Q)\to Q )\to(P\to Q)}$$

Else there is the double negation path.

$$\fitch{}{\fitch{(P\land\lnot Q)\to Q}{\fitch{P}{\fitch{\lnot Q}{P\wedge\lnot Q\\ Q\\\bot}\\\lnot\lnot Q \\Q }\\P\to Q}\\((P\land\lnot Q)\to Q )\to(P\to Q)}$$

If the falsum symbol is not accepted this can be modified to introduce negation via $X\to\neg Y, X\to Y\vdash \neg X$

$$\fitch{}{\fitch{(P\land\lnot Q)\to Q}{\fitch{P}{\fitch{\lnot Q}{\neg Q}\\\lnot Q\to\lnot Q\\\fitch{\lnot Q}{P\wedge\lnot Q\\ Q}\\\neg Q\to Q\\\lnot\lnot Q \\Q }\\P\to Q}\\((P\land\lnot Q)\to Q )\to(P\to Q)}$$

I'll leave the justifications to you.