Prove a finite-dim subspace of a Banach space is always topologically complemented. Let $X$ be a Banach space with a finite-dimensional subspace $F$, then we want to show there is a closed subspace $Y$ of $X$ such that $X = Y \oplus F$.
My attempt is to first just consider algebraically complement (i.e. just in the sense of vector space rather than topological vector space) of $F$ and I try to prove the algebraic projection $\pi: X \to F$ is continuous using the closed graph theorem, but I get stuck.
 A: I would be more inclined to use Hahn-Banach.
Let $\{ f_1, \dots, f_k \}$ be a basis for $F$. Define linear transformations $\phi_1, \dots, \phi_k : F \to \mathbb R$ by
$$ \phi_i(\sum_j c_j f_j) = c_i,$$
where the $c_j$'s are coefficients in $\mathbb R$.
By Hahn-Banach, we can extend $\phi_1, \dots, \phi_k$ to continuous linear functionals $\tilde \phi_1, \dots, \tilde \phi_k$ on the whole of $X$. Since $\tilde \phi_1, \dots, \tilde \phi_k$ are continuous, their kernels are closed linear subspaces.
Now define $Y$ to be the intersection of the kernels of $\tilde \phi_1, \dots, \tilde \phi_k$. $Y$ is clearly a linear subspace, and $Y$ is closed because it is the intersection of closed sets.
Finally, we must check that $X = F \oplus Y$. For any given $x \in X$, we can write $x $ in the form $ f + y$ with $f \in F$ and $y \in Y$ by taking $f = \sum_i \tilde \phi_i(x) f_i$ and $y = x - f$. And it is fairly obvious that $F \cap Y = 0$. So $X = F \oplus Y$.

Remark: You mentioned the closed graph theorem. The closed graph theorem tells you this: if you're know that a Banach space $X$ decomposes as a direct sum of closed linear subspaces $Y_1$ and $ Y_2$, then the projection $X \to Y_1$ is a continuous map. Since your present problem is about showing that $X$ is a direct sum in the first place, the closed graph theorem doesn't help!
