Showing $T: X\rightarrow Y$ is a linear map, is one-to-one... Over-thinking question? so my question is as follows:
Suppose that $X$ and $Y$ are normed linear spaces and that $T: X\rightarrow Y$ is a linear map (ie $T(\alpha x_1+\beta x_2) = \alpha T(x_1) + \beta T(x_2) \forall x_1,x_2 \in X$ and all scalars $\alpha$ and $\beta$). Suppose that $T$ maps $X$ onto $Y$ and is isometric (meaning $||Tx|| = ||x||, \forall x\in X$
1) Show that $T$ is one-to-one
2) Show that if $X$ is a Banach space, so is $Y$
So for number one:
Let $x_1, x_2 \in X$ such that $T(x_1) = T(x_2)$.
Then $||T(x_1 - x_2)|| = ||T(x_1) + T(-x_2)|| = ||T(x_1) - T(x_2)|| = ||0|| = 0$
As $T$ is isomorphic, $||T(x)|| = ||x||, \forall x\in X$.
Therefore $||T(x_1 - x_2)|| = ||x_1 - x_2|| = 0$, and by definition of a norm we know that $||a|| = 0$ implies $a=0$, therefore $x_1-x_2=0 \implies x_1 = x_2$.  
As for number two, I have edited based on the feedback below:
As $X$ is a banach space, let ${x_n}$ be a cauchy sequence in $X$. Then we know $x_n \rightarrow x, x\in X$. For each $\epsilon > 0$, as $x_n \rightarrow x$ we can select $N$ such that, $|x_n - x|<\epsilon ,\forall n\geq N$.
Now we know that $||0|| = 0$ and therefore $||0|| = ||x-x|| = ||\lim_{n \to \infty} x_n - x|| = \lim_{n \to \infty} ||x_n - x|| = 0$
As $T$ is isometric, we know that $||T(x)|| = ||x||$, and therefore
$\lim_{n \to \infty} ||x_n - x|| = \lim_{n \to \infty} ||T(x_n - x)||=\lim_{n \to \infty}||T(x_n)|| - ||T(x)||=0$
Therefore $T(x_n) \to T(x)$, and as $T(x_n)$ is a convergent sequence, and therefore a cauchy sequence, and as $T(x_n)\in Y$, all $T(x_n)$ are cauchy sequences that converge, assuming $x_n$ is cauchy.
Next, let us check if these are all of the cauchy sequences in $Y$. Let us assume that $T(x_j)$ is a cauchy sequence, but that $x_j$ is not cauchy. Then
$0 = \lim_{j \to \infty}||T(x_j)-T(x)|| = \lim_{j \to \infty}||x_n - x||$
Therefore $x_n \to x$, and as $x_n$ is a convergent sequence it is cauchy, however we assumed it was not, therefore there are no $T(x_j)$ which are cauchy in which $x_j$ is not also cauchy.
Finally, as every cauchy sequence in $Y$ converges in $Y$, $Y$ is complete $\implies$ $Y$ is a Banach space.
 A: 1) You received some good critique already, here is way to finish it off.
Note that,
$$0 = T(x) - T(y) = T(x - y).$$
Appealing to the non-negativity of a norm, we get $$0 = \|T(x - y)\| = \| x- y\| \implies x = y.$$
2) Some feedback, 
(i) You are trying to prove that $Y$ is complete, so it is good idea to start with a sequence in $Y$ rather than $X$
(ii) I've noticed you are using the $\epsilon$-$N$ argument to prove convergence; I recommend you use a well-known shortcut, $$\lim_{n \to \infty} x_n = 0 \iff \lim_{n \to \infty} \| x_n - x \| = 0.$$
(iii) At the very end (and in one of your comments), you mentioned these "$y_n$s"; recall that $T$ is onto, so that $\text{Im}(T) = Y.$ 
If you put (ii), (iii) and $\|x_n - x \| = \| T(x_n) - T(x) \|$ together, the result should now follow.
Addendum
I have noticed you were still using Cauchy sequences, here are some facts to help you shorten the proof.
(iv) Every normed space is a metric space; because the norm induces the metric.
(v) Every convergent sequence in a metric space is a Cauchy sequence. The converse is of course not true (e.g. discrete metric).
A: For 1), consider the following:
Suppose that $Tx = Ty$.  What can we say about $T(x-y)$?  How can we conclude $x-y = 0$?
For 2), the key is that $|x_n - x| = |T(x_n) - T(x)|$.
A: $(1)\quad$  $Tx=0$ implies $|x|=|Tx|=0$ and therefore $x=0$. It follows that the kernel of $T$ is $\{0\}$, which means that $T$ is injective.
$(2)\quad$ We now know that $X$ and $Y$ are isomorphic as normed vector spaces, and a fortiori as metric spaces. Therefore $Y$ has to be complete if $X$ is. (Remark: There is no need to consider Cauchy sequences again.)
