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I've read that convergence in one topology implies convergence in coarser ones. (I've also read the contrapositive of this statement). While this make some intuitive sense to me, I can't see exactly why.

I also can't find a good proof of this, however. By "good," I mean one that doesn't use the notion of continuity, which is not yet available to me.

How does one prove this without referring to continuity? In this way, I'm asking for a more elementary (and hence maybe longer) proof.

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    The difficulty of the proof depends on the definition of "convergence" you choose. For example, if you take the position that $x_n \to x$ in a topological space means that every open set that contains $x$ must contain all but finitely many elements of the sequence, the proposition you want to prove is essentially trivial. – 2017-01-20
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    @mathguy Ah, the definition I had was a bit different but I see why it's equivalent to the definition you just gave. So it should be trivial. Thanks! – 2017-01-20

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$\newcommand{\net}[1]{\langle #1 \rangle}$Let $X$ be a set and $\mathcal T,\mathcal U$ topologies on $X$ with $\mathcal{T} \supseteq \mathcal{U}$. Let $\net{x_j}_{j \in M}$ be a net in $X$, where $M$ is a directed set with pre-order $\lesssim$.

If $\net{x_j}_{j \in M}$ converges to $x \in X$ in $\mathcal{T}$, then $\net{x_j}_{j \in M}$ converges to $x$ in $\mathcal{U}$.

Let $S \in \mathcal{U}$ be a neighbourhood of $x$. Then $S \in \mathcal{T}$. Hence $\exists m \in M\mid j \gtrsim m \implies x_j \in S$ by $x_j \stackrel{\mathcal T}{\to} x$, so $x_j \stackrel{\mathcal U}{\to} x$. $\blacksquare$

This also works if "net" is replaced by "sequence" since every sequence is a net.