In a proof I am reading I saw the statement that the set of projections onto one fixed finite dimensional subspace of a Banach space $X$ is compact. Is this obvious, for some reason I just don't see the proof of this apparently simple fact?
Edit 1: All projections have the same range.
Edit 2: The statement in the paper is that there always exists a projection of the smallest norm on a fixed finite dimensional subspace (the inf is attained), and that it follows from a "compactness argument". I wrongly assumed that the set of projections must be compact (thank you Martin). I think what is enough is to consider the set of projection with norm smaller than dimension of the finite dimensional subspace. Is the set compact in this case? If it is, it will prove the claim, as the continuous function $P\to\|P\|$ would attain the minimum on this set.
In general, the set of projections onto a fixed subspace is not bounded, even if $X$ is finite-dimensional.
For instance let $X=\mathbb C^2$, and $$ P_n=\begin{bmatrix}1&n\\0&0\end{bmatrix}. $$ These are all projections onto the subspace $\mathbb C\oplus 0$, and for any norm we have $$ \|P_n-P_0\|=n\,\left\|\begin{bmatrix}0&1\\0&0\end{bmatrix}\right\|. $$ So the diameter of the set of projections onto $\mathbb C\oplus 0$ is infinite.
When $\dim X=\infty$, a bounded set of projections onto a finite-dimensional subspace may still fail to be compact. Let $X=\ell^2 (\mathbb N) $. In matrix unit notation, let $P_n=E_{11}+E_{1n}$ (the first row of $P_n$ is $1\ 0\cdots\ 0\ 1\ 0\ \cdots$ and the rest is zero). Then $P_n$ projects onto the first coordinate, $\|P_n\|=\sqrt2$, and $$ \|P_n-P_m\|=\|E_{1n}-E_{1m}\|=\sqrt2. $$