Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero?

An easy way to visualize why this can't be true is to try putting some points on a number line.

Start with 1 point in [0, 1):

2 points in [1, 2):

And so on:

Now you have a sequence that grows to infinity but keeps getting closer together.

No. Just consider the case in which $a_n=1+\frac12+\frac13+\cdots+\frac1n$. Note that then we would have$$\lim_{n\to\infty}a_{n+1}-a_n=\lim_{n\to\infty}\frac1{n+1}=0.$$

Any increasing sequence $\{a_n\}_{n\geq 1}$ has limit in $\mathbb{R}\cup\{+\infty\}$. It is $\sup_{n\geq 1} a_n$. Such $\sup$ or supremum can be a finite number or $+\infty$ (even if we know that $a_{n+1}-a_n\to 0$).

An example with a finite limit is $a_n=1-1/n\to 1$ and $a_{n+1}-a_n=\frac{1}{n(n+1)}\to 0$.

On the other hand $a_n=\sqrt{n}\to +\infty$ and $a_{n+1}-a_n=\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\to 0$.

So, the answer is NO, the condition $a_{n+1}-a_n\to 0$ is not sufficient for an increasing sequence $\{a_n\}_{n\geq 1}$ to have a FINITE limit.