Let $f : (X ,T) \to (Y,T')$ be continuous and let $T \subset T_1$ and $T_1' \subset T'$. Show that $f : (X,T_1) \to (Y,T_1')$ is continuous. 
Let $f : (X ,T) \to (Y,T')$ be continuous and let $T \subset T_1$ and $T_1' \subset T'$. Show that $f : (X,T_1) \to (Y,T_1')$ is continuous.

Since $f$ is continuous on $(X ,T) \to (Y,T')$ we have that for every $x \in X$ there is an open subset $U_x$ of $X$ such that for every open set $V_{f(x)} \subset Y$ the preimage $f^{-1}(V_{f(x)}) \subset U_x$ is open.
Now since $T_1' \subset T'$ we can pick $V_{f(x)}$ to be in $T_1'$ and now by continuity there is an open subset $U_x \in T$ such that $f^{-1}(V_{f(x)}) \subset U_x$ is open, but $T \subset T_1$ and so $U_x \in T_1$ which would finish the proof? I'm getting a bit confused with the topologies here. Is this correct?

It’s just definitions and applying the inclusions.
If $O$ is open in $T_1’$ it’s open in $T’$ so $f^{-1}[O]$ is in $T$ so in $T_1$ too.
So $f$ is also continuous wrt the new topologies: fewer open sets to test in the codomain and more than enough open sets in the domain…
