Natural deduction proof for $(P\to\lnot Q)\to(\lnot P \lor\lnot Q)$ I can't find the proof for  $(P\to\lnot Q)\to(\lnot P \lor\lnot Q)$ in natural deduction. 
I've tried using the introduction of $\to$ law by assuming $P\to\lnot Q$ and $P$ but can't seem to close my assumption of $P$.
 A: Assuming $P\to \lnot Q$ and the law of excluded middle (so $P\vee \lnot P$) will do it in classical logic.
Hint: The Constructive Dilemna.
$$(A\vee B), (A\to C), (B\to D)~\vdash~ (C\vee D)$$
A: Assume $P\implies \neg Q$ furthermore we know that $P\lor\neg P$ is tautology we may then  argue from cases as follows
Case-1: $P$ is true, then using this together with $P\implies \neg Q$ and applying  modus ponens we can conclude $\neg Q$ and by addition $\neg P\lor \neg Q$.
Case-2: $P$ is false, then from addition we have $\neg P\lor\neg Q$.
$\blacksquare$
A: Following the hint by Luca Bressan, a derivation in natural deduction of $(P \to \lnot Q) \to (\lnot P \lor \lnot Q)$ without any undischarged assumption ($[A]^*$ stands for an assumption $A$ discharged by the rule $*$) is:
\begin{align}
\dfrac{\dfrac{\dfrac{\vdots}{\lnot P \lor P} \qquad \dfrac{[\lnot P]^*}{\lnot P \lor \lnot Q}\lor_{\mathrm{i}_1} \qquad \dfrac{\dfrac{[P \to \lnot Q]^\# \qquad [P]^*}{\lnot Q}\to_\mathrm{e}}{\lnot P \lor \lnot Q}\lor_{\mathrm{i}_2}}{\lnot P \lor \lnot Q}\lor_\mathrm{e}^*}{(P \to \lnot Q) \to (\lnot P \lor \lnot Q)}\to_\mathrm{i}^\#
\end{align}
where $\lnot P \lor P$ can be proved as follows (using reductio ad absurdum, which is valid in classical logic, not in intuitionistic logic):
\begin{align}
\dfrac{\dfrac{[\lnot(\lnot P \lor P)]^* \qquad \dfrac{\dfrac{\dfrac{[\lnot(\lnot P \lor P)]^* \qquad \dfrac{[P]^\#}{\lnot P \lor P}\lor_{\mathrm{i}_2}}{\bot}\lnot_\mathrm{e}}{\lnot P}\lnot_\mathrm{i}^\#}{\lnot P \lor P}\lor_{\mathrm{i}_1}}{\bot}\lnot_\mathrm{e}}{\lnot P \lor P}\mathsf{raa}^*
\end{align}
A: If you don;t have the Law of Excluded Middle readily available to you (i.e. if you first need to prove it doing a proof by contradiction), then I would suggest saving yourself a bit of work and just doing a proof by contradiction directly on your goal $\neg P \lor \neg Q$.
Here is a Fitch-style proof that does this:
$\def\fitch#1#2{\begin{array}{|l}#1 \\ \hline #2\end{array}}$ 
$\fitch{
1. P \rightarrow \neg Q \qquad \quad Assumption}{
\fitch{
2. \neg(\neg P \lor \neg Q) \quad Assumption}{
\fitch{
3. \neg P \qquad \quad \quad Assumption}{
4. \neg P \lor \neg Q \quad \lor \ Intro \ 3\\
5. \bot \qquad \qquad \top \ Intro \ 2,4
} \\
6. \neg \neg P \qquad \quad \neg \ Intro \ 3-5\\
7. P \qquad \qquad \neg \ Elim 6\\
8. \neg Q \qquad \quad \rightarrow \ Elim \ 1,7\\
9. \neg P \lor \neg Q \quad \lor \ Intro \ 8\\
10. \bot \qquad \qquad \top \ Intro \ 2,4}\\
11. \neg \neg (\neg P \lor \neg Q) \quad \neg \ Intro \ 2-10\\
12. \neg P \lor \neg Q \qquad \neg \ Elim \ 11 
}$
