Recall the definition of convergence:
A sequence $\{x_n\}\in X$ converges to $x \in X$ iff for any open neighbourhood $U \ni x$ there exist an $N$ such that $x_i \in U$ for all $i > N$.
If a sequence entirely contained in $[0, 1]$ is to converge toward $z$, then for any neighbourhood of $z$, we must have that eventually the sequence is contained in that neighbourhood. But this isn't the case. I can construct a neighbourhood for which this fails. If the sequence is contained in $[0, 1]$, then it has an infinite subsequence whose complement $U \subset [0, 1]$ is dense and open, and thus $\{z\}\cup U$ is an open neighbourhood of $z$, and the sequence will not eventually be contained there. Therefore the sequence does not converge toward $z$.
As for why sequences in $X$ have unique limits, we can say, with a very similar argument to the above, that if a sequence is to converge to $z$, it must eventually be constant and equal to $z$ (otherwise you can find a neighbourhood around $z$ which the sequence will leave at arbitrarily high indexes). If it does not converge to $z$, but some other number $x \in [0, 1]$, then there is a neighbourhood around $X$ that does not contain $z$. Once the sequence is contained in that neighbourhood (which will happen eventually, by the definition of convergence), then we can pretend $z$ doesn't even exist, and the result follows from the fact that limits are unique on the number line.
