Question about proof that finite-dimensional subspaces of normed vector spaces are direct summands

I am reading a proof that finite-dimensional subspaces of normed vector spaces have closed direct sum complements. This is the proof:

Let $\{e_1, ..., e_n\}$ be a basis for $\mathcal M$. Every $x \in \mathcal M$ has then a unique representation $$x = \alpha_1(x)e_1 +...+ \alpha_n(x)e_n.$$ Each $\alpha_i$ is a continuous linear functional on $\mathcal M$ (a linear map from finite dimensional space is always continuous) which extends to a member of $\mathcal X^*$, by the Hahn-Banach theorem ($\mathcal X^*$ is the dual of $\mathcal X$). Let $\mathcal N$ be the intersection of the null spaces of these exten­sions. Then $\mathcal X = \mathcal M\oplus \mathcal N$.

I follow it until the last sentence. Clearly $\mathcal N$ is closed, as it is the intersection of closed sets (the inverse of $0$ under any continuous function, and in particular any linear functional on a vector space, is closed). Also it is easy to see that $\mathcal M \cap \mathcal N= \{0\}$, since if some $m\in\mathcal M$ is in every null space, it must have the zero coefficient for every basis vector, and so it must be zero.

My question: Why do we have $\mathcal X =\mathcal M + \mathcal N$? This seems to only hold if the extension given by Hahn-Banach is zero off of $\mathcal M$, but I see no reason why this should be the case.

Here is how I would do the proof, in a more explicit way:

After you extend all the $\alpha_i$, you can use them to define a bounded projection onto $\mathcal M$: $$ P(x)=\alpha_1(x)e_1+\cdots+\alpha_n(x)e_n. $$ Now you let $\mathcal N=\ker P$. Then any $x\in\mathcal X$ satisfies $$ x=Px+(x-Px)\in\mathcal M+\mathcal N. $$

Note that from this construction it is clear that $\mathcal N=\displaystyle\bigcap_1^n\ker\alpha_j$, so the original answer was right.