Showing that $\log(\log(N+1)) \leq 1+\sum\limits_{p \leq N} \frac{1}{p}$ I can't see how you get this. 
I want to show that 
$$\log(\log(N+1)) \leq \sum_{p \leq N} \frac{1}{p}+1$$
Can't see how it follows from this. So you show that 
$$0 \lt -\log(1-x)-x \lt \frac{x^2}{(1-x)}$$. 
Which is fine, but then the lecturer does a jump I don't understand.
He rewrites this $$\sum_{p \leq N} \frac{1}{p} + \sum_{p\leq N} (-\log(1-\frac{1}{p}) - \frac{1}{p})$$
Know that
$$ \sum_{p\leq N} (-\log(1-\frac{1}{p}) - \frac{1}{p}) \leq \sum_{p \leq N} \frac{1}{p(p-1)}$$
Then just claims that this shows 
$$\log(\log(N+1)) \leq \sum_{p \leq N} \frac{1}{p}  +1$$
 A: Let's start with :
$$ \sum_{p\leq N} \left(-\log\left(1-\frac{1}{p}\right) - \frac{1}{p}\right) \leq \sum_{p \leq N} \frac{1}{p(p-1)}$$
The idea is to use the finite version of Euler's product :
$$\prod_{p\le N} \frac1{1-p^{-1}} =\prod_{p\le N} \left(1+\frac{1}{p}+\frac{1}{p^2}+\cdots\right)=\sum_{n\in S_N} \frac1n$$
with $S_N$ the set of integers composed of prime factors $p\le N$
Since this set includes all the integers $n\le N$ we have :
$$\prod_{p\le N} \frac1{1-p^{-1}} \ge H_N$$  
but the harmonic sum $H_N$ is well known to be greater than $\log(N+1)$ (use for example minoration by $\int_1^{N+1} \frac1n dn$).
So that (taking logarithms) we get :
$$\log(\log(N+1)) \le \sum_{p\le N} -\log\left(1-\frac1p\right)$$ 
that we may replace in your inequality.

UPDATE2: Let's handle with more care this replacement to show that indeed : $$\log(\log(N+1)) \leq \sum_{p \leq N} \frac{1}{p}  +1$$
from the convergent : $\ \displaystyle 0 \le \sum_{p\leq N} \left(-\log\left(1-\frac{1}{p}\right) - \frac{1}{p}\right) \leq \sum_{p \leq N} \frac{1}{p(p-1)}$
we get (adding the sum on $\frac1p$) : 
$$ \sum_{p\leq N} \frac{1}{p}\le \sum_{p\leq N} -\log\left(1-\frac{1}{p}\right) \leq \sum_{p \leq N} \frac{1}{p-1}$$
At this point we may use the previous result to get not your inequality but :
$$ \log(\log(N+1)) \le \sum_{p \leq N} \frac{1}{p-1}$$
Let's use the fact that $\frac1{p-1}\le \frac1q$ (if $q$ is the prime before $p$) or $1$ (for $p=2$) to rewrite this as ('shifting' the primes and extracting $1$) :
$$ \log(\log(N+1)) \le 1+\left(\sum_{p \le N} \frac1p\right)-\frac1{p_N}\ ,\  \text{with }p_N\ \text{the largest prime}\ \le N$$
this implies indeed :
$$ \log(\log(N+1)) \le 1+\sum_{p \le N} \frac1p$$
We may replace the $1+$ term by $\frac12+$ by noticing that your initial inequality could have been (adding $2$ at the denominator) : 
$$0 \lt -\log(1-x)-x \lt \frac{x^2}{2(1-x)}$$ 
because the effect of this is to replace the majoration of $\frac1{p(p-1)}+\frac1p=\frac1{p-1}$ by $\frac1{2p(p-1)}+\frac1p=\frac{p+p-1}{2p(p-1)}=\frac12(\frac1{p-1}+\frac1p)$ so that the final inequality is replaced by : 
$$ \log(\log(N+1)) \le \frac12\left(1+\sum_{p \le N} \frac1p+\sum_{p \le N} \frac1p\right)$$

We could continue this way but better results may be proved (see for example the excellent Hardy&Wright 'An introduction to the theory of numbers')
using the function $\displaystyle C(x)=\sum_{p\le x} \frac{\log p}p$ and the equality $\displaystyle \sum_{p\le x}\frac1p=\frac{C(x)}{\log x}+\int_2^x \frac{C(t)}{t\ \log^2(t)}dt\ $ to get :
$$\sum_{p\le x}\frac1p= \log\log x+B_1+o(1)$$ 
with $B_1$ the Merten's constant :
$$B_1=\gamma+\sum_p \log\left(1-\frac1p\right)+\frac1p$$
