Completeness, Compactness, Sequentially Compactness for $X = [0,1] \cap \mathbb{Q}$ $X = [0,1] \cap \mathbb{Q} \subset \mathbb{R}$ a metric space with the metric of $\mathbb{R}$. Show $X$ is not complete, is totally bounded, and is not sequentially compact.

For completeness.
I know that $X$ is dense in $[0,1]$. That is, $cl(X) = [0,1]$. For $x \in X$, then $x \in cl(X)$. $\forall_{n \in \mathbb{N}}$ we know that $B_{\frac{1}{n}}(x)$ must contain some point $x_n \in X$. Then the sequence $\{x_n\} \to x$ . We can check this by: for any $\epsilon > 0$, pick $N > \frac{1}{\epsilon}$, so then for all $n > N$ we have $\frac{1}{n} < \frac{1}{N} < \epsilon$, so $|x - x_n| < \frac{1}{n} < \epsilon$  for all such $n$.
This shows that $x_n$ converges outside of $[0,1] \cap \mathbb{Q}$. Since convergent sequences are Cauchy, this means $X$ is not complete. 
For totally boundedness,
A set is totally bounded iff $\exists$ finite $\epsilon$-net for the set. I want to show it is totally bounded. So I need to construct a finite epsilon-net which covers $X = [0,1] \cap \mathbb{Q}$. The farthest $x$ can be is if $x$ is halfway between those two consecutive members of A, so $x = \frac{(\frac{(k-1)}{n} + \frac{k}{n})}{2}$,  so that gives  $|\frac{k}{n} - x| = |\frac{(k-1)}{n} - x| = \frac{(2k)}{(2n)} - \frac{(2k - 1)}{(2n)} = \frac{1}{(2n)}$. So..we have to choose $n$ s.t. $\epsilon > \frac{1}{2n}$ right?
Then the set $\{0,\frac{1}{n}, \frac{2}{n}, ... , 1\}$ is a epsilon net for $X$. Hence, $X$ is Totally Bounded.
For Sequentially compactness
A set is sequentially compact if every (infinite) sequence has a convergent subsequence. This part of the problem, I am not too sure how to do.

Can anyone look over my first two answers and then give advice on how to show that $X$ is not sequentially compact? 

Since $X$ is a metric space, it suffices to show that $X$ is not compact.
Let
$$A_0 = (-1, \frac{\sqrt 2}{2}),$$
$$A_n = (\frac{\sqrt 2}{2} + \frac{1}{n}, 2).$$
Then, $\bigcup A_n$ is an open cover for $X$ with no finite subcover. Hence, $X$ is not sequentially compact.
