Why is the projection of a closed polytope closed? In general, projection of a closed set into a subspace does not result in a closed set. However, I was able to prove that in $\mathbb{R}^n$, the projection of a closed polytope (intersection of finitely many closed half spaces) onto a 1 dimensional subspace is closed. However the proof involved induction on dimension and using the theory of linear optimization.  

Is there a short topological proof of this fact? And perhaps a generalization that projection onto an m-dimensional subspace preserves closure? Intuitively imagining the shadow of a polytope, I feel it's true.
This is a property of linear maps in $\mathbb{R}^n$, not just projections.

The key result here is that the class of polyhedral sets (intersection of a finite number of closed half-spaces) is the same as the class of finitely generated sets (Rockafellar, "Convex analysis", Theorem 19.1).

To quote Rockafellar, "This classical result is an outstanding example of a fact with is completely obvious to geometric intuition, but which wields important algebraic content and is not trivial to prove".

Suppose $P$ is a polyhedral set, then there are vectors $v_1,...,v_l, d_1,....,d_m$ ($l$ or $m$ may be zero with appropriate adjustments, the $v_i$ are 'points', the $d_j$ are 'directions') such that
$P = \{ \sum_i \lambda_i v_i + \sum_j \mu_j d_j | \sum_i \lambda_i =1, \lambda_i \ge 0, \mu_i \ge 0 \}$.

Suppose $A$ is a linear operator. Then
$AP = \{\sum_i \lambda_i A v_i + \sum_j \mu_j A d_j | \sum_i \lambda_i =1, \lambda_i \ge 0, \mu_i \ge 0 \}$, hence $AP$ is finitely generated, and so is a polyhedral set, hence closed.
Projections are continuous functions, and continuous functions preserve limits. If a set contains all of its limit points, then the image of that set under a continuous function also contains all of its limit points. Not sure off the top of my head what the conditions are for this to hold in a general topological setting, but it sure should hold in Euclidean space.