Showing a linear map is injective if and only if kernel is {$ {0} $}

First suppose $f$ is injective.

Since $f$ is linear, $f(0) = 0$, hence $0 \in \text{ker}(f)$.

But if $x$ is any element of $\text{ker}(f)$, then \begin{align*} &x \in \text{ker}(f)&&\\[4pt] \implies\; &f(x) = 0&&\\[4pt] \implies\; &f(x) = f(0)&&\text{[since $f(0) = 0$]}\\[4pt] \implies\; &x = 0&&\text{[since $f$ is injective]}\\[4pt] \end{align*}

It follows that $\text{ker}(f) = \{0\}$.

Thus, $f$ injective implies $\text{ker}(f) = \{0\}$.

Next, suppose $\text{ker}(f) = \{0\}$. Then \begin{align*} &f(x)=f(y)&&\\[4pt] \implies\; &f(x)-f(y) = 0&&\\[4pt] \implies\; &f(x-y) = 0&&\text{[since $f$ is linear]}\\[4pt] \implies\; &x-y \in \text{ker}(f)&&\\[4pt] \implies\; &x-y = 0&&\text{[since $\text{ker}(f) = \{0\}$]}\\[4pt] \implies\; &x=y&&\\[4pt] \end{align*}

hence $f$ is injective.

Thus, $\text{ker}(f) = \{0\}$ implies $f$ is injective.

Hence, $f$ is injective $\iff \text{ker}(f) = \{0\}$, as was to be shown.

I dont see why this proof is enough, this only says $y-x \in Ker f$

It says much more, since you have the symbol $$ \iff $$ meaning that each side implies the other side. Note that $Ker f=\{0\}$ allows one to conclude.