For $n\ge 1$, define $a_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\log n$. Prove that $\lim_{n\to\infty} a_n$ exists.

Here is a pictorial answer. Some people may like it more than a wordy answer.

Let $$ A_n = \sum_{k=1}^{n-1}\frac{1}{k} - \log n $$ Since $A_n$ differs from the term in the question by $\frac{1}{n}$ and $\frac{1}{n} \to 0$, it suffices to show that $A_n$ converges.

Here we have shaded something with area $A_5$.

The lower graph is $y=1/x$ between $x=1$ and $x=5$.
The horizontal line segments are:
$y=1$ from $x=1$ to $x=2$;
$y=1/2$ from $x=2$ to $x=3$;
$y=1/3$ from $x=3$ to $x=4$;
$y=1/4$ from $x=4$ to $x=5$.
The area beow the graph of $y=1/x$ is $\int_1^5\frac{dx}{x} = \log 5$. The area below the horizontal line segments is $\sum_{k=1}^4\frac{1}{k}$. The shaded area is the difference $A_5$.

Now translate these pieces to the left until they bump the $y$-axis.

The pieces are disjoint, and are contained in the square $[0,1] \times [0,1]$. When we go from $A_n$ to $A_{n+1}$ we add one more of these pieces. So the sequence $A_n$ is increasing, and bounded above by $1$. Therefore $A_n$ converges.

Your proof that $a_n>0, \forall n\in\mathbb{N}$ would be correct, except for this line, which you luckily don't use anyway:

This is false : $\lim_{n\to\infty}\sum_{i=1}^n\frac{1}{i}$ exists (no!) and is equal to $\int_1^{n+1}\frac{dx}{x}=\ln(n+1).$

What you probably meant is "$\sum_{i=1}^n \frac{1}{i} > \ln n$ for all $n\in\mathbb{N}$", let's prove that :

First, notice that $$\ln(n) < \ln (n + 1) = \int_1^{n + 1} \frac{1}{x} dx$$

On the other hand, consider the step function $f : x \mapsto \frac{1}{\lfloor x \rfloor}$ , one has $$\sum_{i=1}^n \frac{1}{i} = \int_1^{n + 1} f(x) dx$$

The rest is left to the reader :)