Why if $X \subset A$ has codimension $1$, the only subspace properly containing $X$ is $A$? I saw the following question a while ago on Math.SE This question.
The answer provided seems to give a satisfactory result, but one thing in the answer I can not quite see. The author of the answer states that

Since $X$ has codimension $1$, the only subspace properly containing $X$ is $A$.

I can not seem to see why this is true. Am I missing something? Can anyone please elaborate as to why this is true?
 A: The codimension of $X$ is the dimension of $A/X$, which is one. The subspaces of $A$ containing $X$ are in correspondence with the subspaces of $A/X$ (see below), of which there are only two: the trivial subspace and all of $A/X$.

Let $S(A/X)$ be the set of all subspaces of the vector space $A/X$ and let $S(A;X)$ be the set of all subspaces of $A$ containing $X$. Define a map $$\phi:S(A;X) \to S(A/X)$$ by $\phi(U) = \{ u+X \mid u \in U\}$.
To see that $\phi$ is injective, assume that $\phi(U) = \phi(V)$. Then if $u \in U$, we have $u+X \in \phi(U) = \phi(V)$, so $u+X = v + X$ for some $v \in V$. Therefore $u-v \in X \subseteq V$, so that $u \in V$. By a symmetrical argument we conclude that $U = V$.
Now we show that $\phi$ is surjective. Let $W$ be any subspace of $A/X$. Let $U = \pi^{-1}(W)$, where $\pi:A \to A/X$ is the canonical projection. This $U$ is a subspace of $A$, and moreover it contains $X$ because $\pi(x) = 0_{A/X} \in W$ for any $x \in X$. In fact, $\phi(U) = W$ because
$$\phi(U) = \{u+X \mid u \in \pi^{-1}(W)\} = \{\pi(u) \mid u \in \pi^{-1}(W)\} = W.$$
