I want to prove the statement of the title that kernels of homomorphisms of an algebra are only its two-sided ideals. If $\mathcal{A}$ is an algebra over $\mathbb{K}$, by a two-sided ideal I mean a subalgebra $\mathcal{B}$ such that $a\mathcal{B} \subset \mathcal{A}$ and $\mathcal{B}a \subset \mathcal{A}$ holds for every $a \in \mathcal{A}$.
My reasoning is the following. Suppose $h$ is a homomorphism of this algebra $\mathcal{A}$, so it preserves the operators $h(ab) = h(a)h(b)$, $h(a+b) = h(a) + h(b)$ and $h(ka) = kh(a)$ with $k \in \mathbb{K}$. Consider $\operatorname{Ker}h$, which is a subalgebra of $\mathcal{A}$. Then, if $a \in \operatorname{Ker}h$, we have $h(ba) = h(b)h(a) = 0 = h(a)h(b) = h(ab)$ so $ba \in \operatorname{Ker}h$ and $ab \in \operatorname{Ker}h$. This proves that the kernel of every homomorphism is a two-sided ideal of $\mathcal{A}$.
Conversely, if $\mathcal{B}$ is a two-sided ideal it is, in particular, a vector subspace of $\mathcal{A}$, so it is the kernel of the projection $\pi: \mathcal{A} \to \mathcal{A}/\mathcal{B}$.
Is my proof correct? Also, does this also imply that every vector subspace of $\mathcal{A}$ is a two-sided ideal too?
Your explanation in the converse direction is not very clear. The point is that if $\mathcal{B}$ is a two sided ideal then $\mathcal{A}/\mathcal{B}$ is not just a vector space, but it is an algebra, with the multiplication defined as $(a+\mathcal{B})(a'+\mathcal{B})=aa'+\mathcal{B}$. In this case the projection is a homomorphism of algebras (and not just of vector spaces), and $\mathcal{B}$ is indeed its kernel.
Of course not every subspace is an ideal. For example, the matrix algebra $M_n(F)$ doesn't have any nontrivial two sided ideals, but if $n\geq 2$ then it has many non trivial subspaces.