Proof that if a linear map $T$ admits only $x=0$ as solution to $T(x)=0$, then $T$ is injective (one to one) Theorem: Let $T:\mathbb{R}^n \rightarrow \mathbb{R}^m$ be a linear transformation. Then if $T($x)=0 has only x=0, $T$ is 1-1.
Proof Attempt (by contradiction):
Suppose not. Then $T($x)=b for more than one x. Let $T$(u),$T$(v) map to b, where u,v are unique.
Since $T$ is linear, $T$(u$-$v) = $T$(u) - $T$(v) = b$-$b = 0.
Notice that u $\not=$ v (since they are unique) and therefore u$-$v$ \not=$0.
But, for $T$ to be 1-1: $T$(u) $-$ $T$(v) = 0 $\implies$ $T$(u) $=$ $T$(v) $\implies$ u $=$ v $\implies$ u$-$v$=$0. But since $T$ is linear, and therefore, $T$(u$-$v) = 0, Contradiction. $T$ cannot be 1-1.
Therefore, x $=$ 0 must be the only solution to $T$(x)=0 for $T$ to be 1-1.
Is this viable?
 A: This is not yet a valid proof. You were supposed to show that if there are distinct $u,v$ such that $T(u)=T(v)$, then there is a $w\not =0$ such that $T(w)=0$. [You do note that $T(u-v)=0$ and that $u-v$ is nonzero, but then you just kept on going. Part of a valid proof is concluding properly.]
You were close though: If $T$ is not one-to-one, then by definition there exists two distinct vectors, $u$ and $v$, such that $T(u)=T(v)$. However: $$T(u)=T(v); \ u \not = v;$$ $$\implies T(u)-T(v) =0; \ u \not = v$$ $$\implies T(u-v)=0; \ u-v \not = 0.$$ Thus, if $T$ is not one-to-one, then there is a $w \not = 0$ such that $T(w)=0$, namely $w=u-v$. Thus as the contrapositive of this is equivalent, it is also true: If $T(w)=0$ only for $w=0$, then $T$ is on-to-one. [Edited to start from the beginning of the proof to the end.]
A: Here is a proof in both directions:
(1)  Assume that $T(x)=0 \implies x=0$, then \begin{align}
 T(a)=T(b) & \implies T(a-b)=0 \\
           & \implies a-b=0 \\
           & \implies a=b \\
\end{align}
Therefore $T$ is one-one.
(I think this is what is asked to  be proved.)
(2) Conversely assume that $T$ is one-one, then
\begin{align}
T(x)=0 & \implies T(x)=T(0) \\
       & \implies x=0 \;\;\;\; \mathrm{ (because \;T\;  is \; one-one) }\\
\end{align}
A: 
… where $\mathbf{u}$ and $\mathbf{v}$ are unique.

Sorry, but this means nothing. What you want to say is that $\mathbf{u}$ and $\mathbf{v}$ are different. So $\mathbf{u}\ne\mathbf{v}$ by assumption (and by way of contradiction) and does not follow from anything else.
After this, you lose yourself, I'm afraid.
From the assumption that $T(\mathbf{u})=\mathbf{b}=T(\mathbf{v})$ you can deduce $T(\mathbf{u})=T(\mathbf{v})$. Then
$T(\mathbf{u}-\mathbf{v})=\mathbf{0}$. By the assumption on $T$, this implies $\mathbf{u}-\mathbf{v}=\mathbf{0}$, which contradicts $\mathbf{u}\ne\mathbf{v}$.
Actually, you don't need contradiction.
Suppose $T(\mathbf{u})=T(\mathbf{v})$. Then $T(\mathbf{u}-\mathbf{v})=\mathbf{0}$ by linearity and so, by assumption, $\mathbf{u}-\mathbf{v}=\mathbf{0}$. Therefore $\mathbf{u}=\mathbf{v}$. This proves that $T$ is 1-1.
A: In fact you can apply a stronger theorem: Let $f:V\rightarrow W$ be a linear mapping, then $f(u)=f(v)\Longleftrightarrow u-v\in kerf$
sketch of proof: for $\Rightarrow$, $f(u-v)=f(u)-f(v)=0$
for$\Leftarrow$, $f(u)=f(v+u-v)=f(v)+f(u-v)=f(v)$ $\square$
Given this lemma, now if $f$ is injective, $u-v\in kerf\Rightarrow f(u)=f(v)\Rightarrow u=v$, in other words $u-v=0\Rightarrow kerf=\{0\}$
