If ${g_n g_{n+1}^{-1}}$ were eventually constant (let's say $g_n g_{n+1}^{-1} = g_m g_{m+1}^{-1}$ for all $n,m \geq N$, where $N$ is some fixed natural number), then the limit of the sequence (which is unique because metric topologies are Hausdorff) would be $g_N g_{N+1}^{-1}$. At the same time, the limit of the sequence is $1$, so $g_n g_{n+1}^{-1} = 1$ for all $n \geq N$. This means that $g_n = g_{n+1}$ for all $n \geq N$, and thus $g_n \to g_N$. Since $g_n \to g$, w have $g = g_N$. This is a contradiction because $g \in G - \Gamma$ but $g_N \in \Gamma$.
Edit: to answer your second question, the metric is actually used in this argument to get the sequence $g_n$. But you're right that Hausdorff is all that you really need: suppose $G$ is Hausdorff and $\Gamma$ is a non-closed discrete subgroup. Then there is some limit point $g$ of $\Gamma$ such that $g \notin \Gamma$. Take a sequence $g_n \to g$ with each $g_n \in \Gamma$ and use the same argument as in this proof. It's perhaps a little harder to see that $\{g_n g_{n+1}^{-1}\}$ must be convergent, but it's still doable:
Let $U$ be an open set containing $1$. Let $V = \{(x,y) \in G \times G : xy^{-1} \in U\}$. $V$ is also open because $G$ is a topological group. The sequence $\{(g_n, g_{n+1})\}$ converges to $(1,1)$, so there is some $N \in \mathbb{N}$ such that $(g_n, g_{n+1}) \in V$ for all $n \geq N$. This means that $g_n g_{n+1}^{-1} \in U$ for all $n \geq N$, and since $U$ was an arbitrary open neighborhood of $1$, we have that $g_n g_{n+1}^{-1} \to 1$.
Edit 2 The first edit is wrong – the result is still true in general when $G$ is Hausdorff, but my proof assumes that "$\Gamma$ is not closed" implies "there exists a sequence in $G - \Gamma$ which converges to a point in $\Gamma$", which is not true in all topological spaces! We need to use a different argument; thanks to @HennoBrandsma for linking to this relatively simple approach.
Here is a solution. Although I can't imagine that it is what the instructor had in mind for the exercise, it does completely destroy the problem.
It's a standard fact that any space $X$ can be embedded as a closed subspace of a contractible space. The usual construction is to use the cone $CX=X\times[0,1]/X\times\{1\}$ and embed $X$ as $X\times\{0\}$. While this is satsifactory for many applications, it has many faults. For one thing it does not preserve subspaces. Another is that it does not preserve separation properties past $T_2$. More relevantly for us is that $CX$ need not be locally contractible, and that $CX$ carries no group structure.
Here is a construction which remedies some of these defects. In particular it will embed any (Hausdorff) topological group into a contractible, locally contractible (Hausdorff) topological group. Note that every contractible space is path-connected. In the Hausdorff case we can replace 'path' everywhere with 'arc' (this is really a consequence of the Hahn–Mazurkiewicz Theorem, although see here for some details).
Let $X$ be a space. A right-continuous step function in $X$ is a map $f:[0,1)\rightarrow X$ for which there is a finite partition $t_0=0<t_1<\dots<t_n<1=t_{n+1}$ of $[0,1)$ such that $f$ is constant on $[t_i,t_{i+1})$ for each $i=0,\dots,n$. Let $EX$ denote the set of all right-continuous step functions $[0,1)\rightarrow X$.
For $t_0<t_1\in [0,1]$, $U\subset X$ open and $\epsilon>0$ let $N_\epsilon(t_0<t_1,U)\subseteq EX$ be the set of all $f\in EX$ with the property that the set $\{t\in [t_0,t_1)\mid f(t)\not\in U\}$ has Lebesgue measure $<\epsilon$. Topologise $EX$ by giving it the subbase $\{N_\epsilon(t_0<t_1,U)\mid t_0<t_1\in[0,1),\;U\subseteq X\;\text{open},\;\epsilon>0\}$. Note that a function $f\in EX$ has a neighbourhood subbase consisting of those sets $N_\epsilon(t_0<t_1,U)$ where $f$ is constant on $[t_0,t_1)$ and $f(t_0)\in U$.
There is a function $i_X:X\rightarrow EX$ which sends a point $x\in X$ to the constant step function function at $x$.
Let $X$ be a nonempty space. Then $EX$ is contractible and locally contractible. The map $i_X:X\rightarrow EX$ is an embedding, which is closed if $X$ is Hausdorff. If $X$ is $T_i$ for some $i\in\{0,1,2,3\frac{1}{2}\}$, then $EX$ if $T_i$. If $X$ is completely regular, then $EX$ is completely regular. If $X$ is first-countable/second-countable/separable/metrisable, then $EX$ is first-countable/separable/metrisable.
It's worth recording that $EX$ does not have all good properties that $X$ may have. The space $EX$ need not be normal, paracompact, locally compact, completely metrisable, or finite-dimensional, even when $X$ is.
The construction is functorial. A map $\alpha:X\rightarrow Y$ induces $E\alpha:EX\rightarrow EY$, $f\mapsto \alpha\circ f$, which is continuous and satisfies $E\alpha \circ i_X=i_Y\circ\alpha$. It can be shown that if $\alpha$ is an embedding, then so is $E\alpha$.
For nonempty spaces $X,Y$, the natural map $E(X\times Y)\rightarrow EX\times EY$ is a homeomorphism.
Now let $G$ be a topological group. The multiplication $m:G\times G\rightarrow G$ induces a map
$$\mu:EG\times EG\cong E(G\times G)\xrightarrow{Em}EG$$
and similarly the inversion $G\rightarrow G$, $g\mapsto g^{-1}$ gives rise to $\iota:EG\rightarrow EG$. It is easy to use the functorality to see that $\mu$ furnishes $EG$ with a continuous multiplication for which $\iota$ is a continuous inverse. Moreover, with these definitions the map $i_G:G\rightarrow EG$ is a homomorphism.
In summary;
Let $G$ be a topological group. Then $G$ embeds into a contractible, locally contractible topological group $EG$. If $G$ is Hausdorff, then so is $EG$, and moreover $G$ is closed in $EG$ in this case. If $G$ is abelian/divisible/torsion/torsion-free, then so is $EG$.
So, as promised, the exercise has been completely demolished. To keep the length somewhat sane I haven't included too many details. If you would like to follow them up, the construction is due to R. Brown and S. Morris in the joint paper
Embeddings in contractible or compact objects*, Coll. Math. 38 (1978), 213-222. Some further details are found in a follow up paper of the second author. (Edit: The topology I give above differs from that given in the reference. While I checked many of the details with my description, I have quoted many unchecked. You should believe the Brown-Morris paper before me.)
The construction has applications in topology, but was actually inspired by the group-theoretic problem. I believe it was in fact S. Hartman and J. Mycielski's paper On the imbedding of topological groups into connected topological groups Coll. Math. 5 (1958) 167-169, which inspired the construction.
In fact, given the name of the paper, I'd recommend you might like to start with this earlier paper :P.
Best Answer
Let $H$ be a discrete subgroup of a Hausdorff group $G$. Then, of course, $H=\bigcup_{h\in H}\{h\}$. For each $h\in G$, the sets $\{h\}$ is closed, and the family of closed sets $\mathcal F=\{\{h\}:h\in H\}$ is locally finite: it follows that the union is also closed.
Locally finite means: each point of $G$ has a neighborhood $U$ such that $U$ intersects non-trivially finitely many elements of $\mathcal F$.
Later. I guess a hint was not all that was asked for... Of course, we now have to show that $\mathcal F$ is locally finite... Suppose, to reach a contradiction, that there is a $p\in G$ such that every neighborhood of $p$ contains infinitely many elements of $H$. As $H$ is discrete, there is an open subset $U\subseteq G$ such that $U\cap H=\{e\}$; since $G$ is a topological group, there exists an open set $V$ such that $e\in V$ and $V^{-1}V\subseteq U.$ Now $pV$ is an open neighborhood of $p$ so the hypothesis implies that there exist distinct $h_1$, $h_2\in H$ such that $h_1$, $h_2\in pV$. It follows that $h_1^{-1}h_2\in V^{-1}V\subseteq U$ so that $e\neq h_1^{-1}h_2\in U\cap H$. This is absurd.