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Looks pretty good. Two minor points.

 Quote by moweee Hello all. I have to present a proof to our Intro to Topology class and I just wanted to make sure I did it right (before I look like a fool up there). Proposition Let c be in ℝ such that c≠0. Prove that if {an} converges to a in the standard topology, denoted by τs, then {can} converges to ca in the standard topology on ℝ. Proof Let c $\in$ ℝ such that c≠0. Suppose {an} converges to a in the standard topology, denoted by τs. Let V $\in$ τs with ca in V. Since V in τs, there exists an interval (p,q) with ca $\in$ (p,q) and (p,q) $\subseteq$ V. Thus, p < ca < q, which implies p/c < a < q/c.
Beware that c might be negative.

 Note that p/c < (p/c + a)/2 < a < (q/c + a)/2 < q/c
I don't really see the point of this line. Why is this necessary??

 Thus, (p/c , q/c) $\in$ τs such that a $\in$ (p/c, q/c). Since, by our assumption, {an} converges to a in the standard topology, there exists m $\in$ N such that an $\in$ (p/c,q/c) for all n ≥ m. Hence, p/c < an < q/c. Hence, p < can< q for all n≥m. Since can $\in$ (p,q) and (p,q) $\subseteq$ V, can $\in$V for all n ≥ m. Therefore, {can} converges to ca in the standard topology on ℝ.